Trigonometric Ratios & Identities

We start with the equation $8 \cos \theta + 15 \sin \theta = 15$. We need to find $15 \cos \theta - 8 \sin \theta$, which we will denote as $t$.

First, consider squaring both sides of the given expression for clarity:

Start with $(8 \cos \theta + 15 \sin \theta)^2 = 15^2$:

$ 64 \cos^2 \theta + 225 \sin^2 \theta + 2 \cdot 8 \cdot 15 \cos \theta \sin \theta = 225. $

Simplify to:

$ 64 \cos^2 \theta + 225 \sin^2 \theta + 240 \cos \theta \sin \theta = 225. $

For $t = 15 \cos \theta - 8 \sin \theta$, square both sides:

$ (15 \cos \theta - 8 \sin \theta)^2 = t^2. $

Expanding this gives:

$ 225 \cos^2 \theta + 64 \sin^2 \theta - 240 \cos \theta \sin \theta = t^2. $

Now, add the two squared expressions:

$ 64 \cos^2 \theta + 225 \sin^2 \theta + 240 \cos \theta \sin \theta + 225 \cos^2 \theta + 64 \sin^2 \theta - 240 \cos \theta \sin \theta = 225 + t^2. $

Simplify the combined expression:

$ 289 (\cos^2 \theta + \sin^2 \theta) = 225 + t^2. $

Since $\cos^2 \theta + \sin^2 \theta = 1$, it follows that:

$ 289 = 225 + t^2. $

Solve for $t^2$:

$ 289 - 225 = t^2, $

$ 64 = t^2, $

$ t = 8. $

Thus, $15 \cos \theta - 8 \sin \theta = 8$.

$ \sin 20^{\circ}\left(4+\sec 20^{\circ}\right) $

$ =\sin 20^{\circ}\left(\frac{4 \cos 20^{\circ}+1}{\cos 20^{\circ}}\right) $

$ \begin{aligned} & =\frac{\left[4 \sin 20^{\circ} \cos 20^{\circ}+\sin 20^{\circ}\right]}{\cos 20^{\circ}} \\ & =\frac{\left[2 \sin 40^{\circ}+\sin 20^{\circ}\right]}{\cos 20^{\circ}} \\ & =\frac{2 \sin \left(60^{\circ}-20^{\circ}\right)+\sin 20^{\circ}}{\cos 20^{\circ}} \\ & =\frac{\frac{2 \sqrt{3}}{2} \cos 20^{\circ}-2 \frac{1}{2} \sin 20^{\circ}+\sin 20^{\circ}}{\cos 20^{\circ}} \\ & =\frac{\sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ}}=\sqrt{3} \end{aligned} $

Given that $ \sinh x = \frac{12}{5} $, we denote $ \cosh x = t $. We utilize the identity for hyperbolic functions:

$ \cosh^2 x - \sinh^2 x = 1 $

Substituting the known value, we have:

$ t^2 - \left(\frac{12}{5}\right)^2 = 1 $

$ t^2 - \frac{144}{25} = 1 $

Solving for $ t^2 $, we get:

$ t^2 = \frac{169}{25} $

Thus, $ t = \frac{13}{5} $.

Now, we need to find $ \sinh 3x + \cosh 3x $. We use the identities for triple angles:

$ \sinh 3x = 3 \sinh x + 4 \sinh^3 x $

$ \cosh 3x = 4 \cosh^3 x - 3 \cosh x $

So,

$ \sinh 3x + \cosh 3x = 3\sinh x + 4\sinh^3 x + 4\cosh^3 x - 3\cosh x $

We substitute the values:

$ = 3\left(\frac{12}{5} - \frac{13}{5}\right) + 4\left(\frac{12}{5} + \frac{13}{5}\right)\left(\frac{144}{25} + \frac{169}{25} - \frac{156}{25}\right) $

$ = 3\left(\frac{12}{5} - \frac{13}{5}\right) + 4\left(\frac{25}{5}\right)\left(\frac{157}{25}\right) $

$ = \frac{-3}{5} + 4 \cdot \frac{157}{25} $

$ = \frac{-3}{5} + \frac{628}{25} $

$ = \frac{-3}{5} + \frac{628}{25} = \frac{-3}{5} + \frac{2512}{125} $

$ = \frac{-75 + 2512}{125} = \frac{2437}{125} $

Thus, the calculated value is:

$ = 125 $

Given,

$ \because \tan A=-\frac{60}{11} $, A lies in 2nd quadrant

$ \begin{array}{l} \sec B=\frac{41}{9}, B \text { lies in } 4 \text { th quadrant } \\\\ \begin{aligned} \operatorname{cosec} A & =\sqrt{1+\cot ^{2} A}=\sqrt{1+\frac{121}{3600}} \\\\ & =\sqrt{\frac{3721}{3600}}=\frac{61}{60} \end{aligned} \\\\ \begin{array}{l} \cot B=\frac{1}{\tan B}=\frac{1}{-\sqrt{\sec ^{2} B-1}} \\\\ =\frac{1}{-\sqrt{\frac{1681}{81}-1}=\frac{-9}{40}} \\\\ \begin{aligned} \operatorname{cosec} A & +\cot B=\frac{61}{60}-\frac{9}{40} \\\\ \quad & =\frac{122-27}{120}=\frac{95}{120}=\frac{19}{24} \end{aligned} \\\\ \therefore \quad 24 K=19 \end{array} \end{array} $

Given:

$ \cos^2 84^\circ + \sin^2 126^\circ - \sin 84^\circ \cos 126^\circ = K $

We can simplify this as follows:

$\sin^2 126^\circ = \sin^2 (90^\circ + 36^\circ) = \cos^2 36^\circ$ by using the identity $\sin(90^\circ + \theta) = \cos \theta$.

$-\sin 84^\circ \cos 126^\circ = \cos (90^\circ - 84^\circ) \cos (90^\circ + 36^\circ)$.

Substituting these identities, we get:

$ K = \cos^2 84^\circ + \cos^2 36^\circ - \sin 84^\circ \cos 126^\circ $

Next, using the identity for product-to-sum, $\sin \theta \cos \phi = \frac{1}{2} [\sin (\theta + \phi) + \sin (\theta - \phi)]$, substitute values as needed, ultimately simplifying:

$ K = \cos^2 84^\circ - \sin^2 36^\circ + 1 + \frac{3}{4} - \sin 24^\circ $

Continuing to simplify using trigonometric identities:

$ K = \cos 120^\circ \cos 48^\circ + \frac{7}{4} - \sin^2 24^\circ $

Now solve for $K$:

$ K = -\frac{1}{2} (1 - 2 \sin^2 24^\circ) + \frac{7}{4} - \sin^2 24^\circ $

This results in:

$ K = -\frac{1}{2} + \frac{7}{4} = \frac{5}{4} $

Now, for the expression involving $\tan A$:

$ \tan A + \cot A = 2K $

Substitute $K = \frac{5}{4}$:

$ \tan A + \frac{1}{\tan A} = \frac{5}{2} $

Multiply through by $\tan A$ to eliminate the fraction:

$ 2\tan^2 A - 5\tan A + 2 = 0 $

Factor the quadratic equation:

$ (2\tan A - 1)(\tan A - 2) = 0 $

Thus, the possible values for $\tan A$ are:

$ \tan A = \frac{1}{2}, 2 $

Given, $ f(x)=\sec x $

$ f^{\prime}(x)=\sec x \tan x $

take $ a=60^{\circ}=\frac{\pi}{3} $ and $ h=1^{\circ}=0.0174^{\circ} $ $ f(a)=\sec \left(\frac{\pi}{3}\right)=2 $

$ \begin{array}{l} \Rightarrow f^{\prime}(a)=\sec \left(\frac{\pi}{3}\right) \tan \left(\frac{\pi}{3}\right)=2 \times \sqrt{3} \\\\ =f(a-h) \approx f(a)-h f^{\prime}(a) \\\\ \approx 2-0.0174 \times 2 \sqrt{3} \\\\ \approx 2(1-0.0174 \times 1.732)=1.9397 \end{array} $

We have, $f(x)=3 \sin ^{12} x+4 \cos ^{10} x$

Period of $3 \sin ^{12} x$ is $\pi$.

Period of $4 \cos ^{16} x$ is $\pi$.

$\therefore$ Period of $f(x)=3 \sin ^{12} x+4 \cos ^{16} x$ is $\pi$

Maximum of $f(x)=$ Maximum of $f(x)$ in $[0, \pi]$

$ f(x)=3 \sin ^{12} x+4 \cos ^{16} x $

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & f^{\prime}(x)=36 \sin ^{11} x \cos x+64 \cos ^{15} x(-\sin x) \\\\ & \Rightarrow f^{\prime}(x)=4 \sin x \cos x\left(9 \sin ^{10} x-16 \cos ^{14} x\right) \\\\ & \Rightarrow f^{\prime}(x)=2 \sin 2 x\left(9 \sin ^{10} x-16 \cos ^{14} x\right) \end{aligned} $

On putting $f^{\prime}(x)=0$, we get

$ \begin{aligned} & 2 \sin 2 x\left(9 \sin ^{10} x-16 \cos ^{14} x\right)=0 \\\\ & \Rightarrow \quad 2 \sin 2 x=0 \Rightarrow x=0, \frac{\pi}{2}, \pi \end{aligned} $

Clearly, the maximum value of $f(x)$ is 4 .

We have,

$ \begin{aligned} & \quad \cos x+\cos y=\frac{2}{3} \text { and } \sin x-\sin y=\frac{3}{4} \\\\ & \therefore 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{2}{3} \quad \ldots \text { (i) } \\\\ & \text { and } 2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)=\frac{3}{4} \quad \ldots \text { (ii) } \end{aligned} $

Using Eqs. (i) and (ii), we get

$ \begin{aligned} & \tan \left(\frac{x-y}{2}\right)=\frac{9}{8} \\\\ & \because \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \\\\ & \therefore \tan (x-y)=\frac{2\left(\frac{9}{8}\right)}{1-\left(\frac{9}{8}\right)^2}=-\frac{144}{17} \end{aligned} $

Now, $\sin (x-y)=\frac{144}{145}$ and $\cos (x-y)=-\frac{17}{145}$

$\begin{aligned} \therefore \sin (x-y)+\cos (x-y) & =\frac{144}{145}+\left(-\frac{17}{145}\right) \\\\ & =\frac{127}{145}\end{aligned}$

We have

$ \begin{aligned} & \tan 2 A=-\frac{4}{3} \\\\ & \tan ^2 2 A=\sec ^2 2 A-1 \\\\ & \frac{16}{9}+1=\sec ^2 2 A \Rightarrow \sec ^2 2 A=\frac{25}{9} \\\\ & \cos ^2 2 A=\frac{9}{25} \end{aligned} $

$\begin{aligned} & \Rightarrow \quad \cos 2 A=\frac{3}{5} \quad(\because \tan A, \tan 2 A<0) \\\\ & \begin{aligned} \because \cos 6 A & =\cos 3(2 A)=4 \cos ^3 2 A-3 \cos 2 A \\\\ & =4\left(\frac{3}{5}\right)^3-3\left(\frac{3}{5}\right) \\\\ & =\frac{3}{5}\left(\frac{36}{25}-3\right) \\\\ & =\frac{3}{5} \times\left(\frac{-39}{25}\right)=\frac{-117}{125}\end{aligned}\end{aligned}$

We have,

$ \begin{aligned} & m \cos (\alpha+\beta)-n \cos (\alpha+\beta) \\\\ & =m \cos (\alpha-\beta)+n \cos (\alpha+\beta) \\\\ & m[(\cos (\alpha+\beta)-\cos (\alpha-\beta)]=n \\\\ & {[\cos (\alpha+\beta)+\cos (\alpha-\beta)]} \end{aligned} $

$\begin{aligned} & m\left[\begin{array}{l}-2 \sin \left(\frac{\alpha+\beta+\alpha-\beta}{2}\right) \\\\ \sin \left(\frac{\alpha+\beta-\alpha+\beta}{2}\right)\end{array}\right] \\\\ & =n\left[\begin{array}{l}2 \cos \left(\frac{\alpha+\beta+\alpha-\beta}{2}\right) \\\\ \cos \left(\frac{\alpha+\beta-\alpha+\beta}{2}\right)\end{array}\right] \\\\ & m[-2 \sin \alpha \sin \beta]=n[2 \cos \alpha \cos \beta] \\\\ & -m \tan \alpha \tan \beta=n \\\\ & \tan \alpha \tan \beta=-\frac{n}{m}\end{aligned}$

$ \begin{aligned} & \text { } \tan ^2 \frac{\pi}{16}+\tan ^2 \frac{2 \pi}{16}+\tan ^2 \frac{3 \pi}{16} \\ & \begin{array}{r} +\tan ^2 \frac{4 \pi}{16}+\tan ^2 \frac{5 \pi}{16}+\tan ^2 \frac{6 \pi}{16} \\ +\tan ^2 \frac{7 \pi}{16}=\left(\sec ^2 \frac{\pi}{16}-1\right)+\left(\sec ^2 \frac{2 \pi}{16}-1\right) \\ +\left(\sec ^2 \frac{3 \pi}{16}-1\right)+1+\tan ^2\left(\frac{\pi}{2}-\frac{3 \pi}{16}\right) \\ +\tan ^2\left(\frac{\pi}{2}-\frac{2 \pi}{16}\right)+\tan ^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right) \end{array} \\ & \begin{array}{r} \left(\sec ^2 \frac{\pi}{16}-1\right)+\left(\sec ^2 \frac{2 \pi}{16}-1\right) +\left(\sec ^2 \frac{3 \pi}{16}-1\right)+1 \end{array} \end{aligned} $

$ \begin{aligned} & +\cot ^2\left(\frac{3 \pi}{16}\right)+\cot ^2\left(\frac{2 \pi}{16}\right)+\cot ^2\left(\frac{\pi}{16}\right) \\ & =\left(\sec ^2 \frac{\pi}{16}-1\right)+\left(\sec ^2 \frac{2 \pi}{16}-1\right) \\ & +\left(\sec ^2 \frac{3 \pi}{16}-1\right)+1+\left(\operatorname{cosec}^2 \frac{3 \pi}{16}-1\right) \\ & +\left(\operatorname{cosec}^2 \frac{2 \pi}{16}-1\right)+\left(\operatorname{cosec}^2 \frac{\pi}{16}-1\right) \\ & =\left(\sec ^2 \frac{\pi}{16}+\operatorname{cosec}^2 \frac{\pi}{16}\right) +\left(\sec ^2 \frac{2 \pi}{16}+\operatorname{cosec}^2 \frac{2 \pi}{16}\right) +\left(\sec ^2 \frac{3 \pi}{16}+\operatorname{cosec}^2 \frac{3 \pi}{16}\right)-6+1 \end{aligned} $

$ \begin{aligned} & =\frac{1}{\sin ^2 \frac{\pi}{16} \cos ^2 \frac{\pi}{16}}+\frac{1}{\sin ^2 \frac{2 \pi}{16} \cos ^2 \frac{2 \pi}{16}} \\ & +\frac{1}{\sin ^2 \frac{3 \pi}{16} \cos ^2 \frac{3 \pi}{16}}-5 \\ & =\frac{4}{\sin ^2 \frac{\pi}{8}}+\frac{4}{\sin ^2 \frac{\pi}{4}}+\frac{4}{\sin ^2 \frac{3 \pi}{8}}-5 \\ & =4\left(\frac{1}{\sin ^2 \frac{\pi}{8}}+\frac{1}{\cos ^2 \frac{\pi}{8}}\right)+4 \times 2-5 \\ & =4\left(\frac{1}{\sin ^2 \frac{\pi}{8} \cos ^2 \frac{\pi}{8}}\right)+3=\frac{4 \times 4}{\sin ^2 \frac{\pi}{4}}+3 \\ & =2 \times 16+3=35 \end{aligned} $

$ \begin{aligned} & \text { } \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ} +\sin ^2 42^{\circ}+\sin ^2 78^{\circ}+\sin ^2 90^{\circ} \\ & +\sin ^2 96^{\circ}+\sin ^2 102^{\circ}+\sin ^2 138^{\circ} +\sin ^2 162^{\circ} \\ & \because \sin ^2 \theta=\sin ^2\left(180^{\circ}-\theta\right) \end{aligned} $

$ \begin{aligned} & \begin{array}{r} \therefore \sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ} +\sin ^2 78^{\circ}+1+\sin ^2 84^{\circ} +\sin ^2 78^{\circ}+\sin ^2 42^{\circ}+\sin ^2 18^{\circ} \end{array} \\ & \begin{array}{r} =2 \sin ^2 18^{\circ}+2 \sin ^2 42^{\circ}+2 \sin ^2 78^{\circ} +\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 84^{\circ}+1 \end{array} \\ & \begin{array}{r} =2 \sin ^2 18^{\circ}+2 \sin ^2 42^{\circ}+2 \cos ^2 12^{\circ} +\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\cos ^2 6^{\circ}+1 \end{array} \\ & \begin{array}{r} 2\left[\cos ^2 12^{\circ}+\sin ^2\left(30^{\circ}-12^{\circ}\right)\right. \left.+\sin ^2\left(30^{\circ}+12^{\circ}\right)\right] \end{array} \\ & +\begin{array}{r} {\left[\cos ^2 6^{\circ}+\sin ^2\left(30^{\circ}+6^{\circ}\right)+\sin ^2\left(30^{\circ}-6^{\circ}\right)\right]} +1 \end{array} \\ & \begin{array}{r} \because \cos ^2 \theta+\sin ^2\left(30^{\circ}+\theta\right)+\sin ^2\left(30^{\circ}-\theta\right)=0 \\ \cos ^2 \theta+\left[\sin 30^{\circ} \cos \theta+\cos 30^{\circ} \sin \theta\right]^2 \\ +\left[\sin 30^{\circ} \cos \theta-\cos 30^{\circ} \sin \theta\right]^2 \end{array} \\ & \text { Using }(a+b)^2+(a-b)^2=2\left(a^2+b^2\right) \end{aligned} $

$ \begin{aligned} & \begin{aligned} \Rightarrow & \cos ^2 \theta+2\left[\sin ^2 30^{\circ}\right. \cos ^2 \theta\left.\quad+\cos ^2 30^{\circ} \sin ^2 \theta\right] \end{aligned} \\ & \Rightarrow \cos ^2 \theta+2\left[\frac{1}{4} \cos ^2 \theta+\frac{3}{4} \sin ^2 \theta\right] \\ & \Rightarrow \cos ^2 \theta+\frac{1}{2} \cos ^2 \theta+\frac{3}{2} \sin ^2 \theta \\ & \Rightarrow 3 / 2 \quad\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]\\ & \end{aligned} $

$ \begin{aligned} &\text { Using this result in Eq. (i), we get }\\ &2\left(\frac{3}{2}\right)+\frac{3}{2}+1=\frac{11}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \frac{\sin A+\sin B+\sin C}{\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}-1} \\ & =\frac{2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)+2 \sin \frac{B}{2} \cos \frac{B}{2}}{\sin ^2 \frac{A}{2}-\cos ^2 \frac{C}{2}+\sin ^2 \frac{B}{2}} \\ & =\frac{2 \cos \frac{B}{2} \cos \frac{A-C}{2}+2 \sin \frac{B}{2} \cos \frac{B}{2}}{-\cos \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)+\sin ^2 \frac{B}{2}} \\ & {\left[\because \sin ^2 A-\cos ^2 B=-\cos (A+B) \cos (A-B)\right]} \\ & \quad=\frac{2 \cos \frac{B}{2}\left[\cos \left(\frac{A-C}{2}\right)+\sin \frac{B}{2}\right]}{-\sin \frac{B}{2}\left[\cos \left(\frac{A-C}{2}\right)+\sin \frac{B}{2}\right]} \\ & \quad=-2 \cot \frac{B}{2} \end{aligned} \end{aligned} $

We start with the given equations:

$ \begin{aligned} \cos \alpha + 4 \cos \beta + 9 \cos \gamma &= 0 \quad ...(i) \\ \sin \alpha + 4 \sin \beta + 9 \sin \gamma &= 0 \quad ...(ii) \end{aligned} $

Combine these equations using the complex form of the equations. Multiply both expressions:

$ \cos \theta + i \sin \theta = e^{i\theta} $

Thus, combining both equations, we have:

$ \begin{aligned} & (\cos \alpha + i \sin \alpha) + 4 (\cos \beta + i \sin \beta) + 9 (\cos \gamma + i \sin \gamma) = 0 \\ & e^{i\alpha} + 4 e^{i\beta} + 9 e^{i\gamma} = 0 \\ & \Rightarrow 1 + 4 e^{i(\beta-\alpha)} = -9 e^{i(\gamma-\alpha)} \end{aligned} $

Substitute using Euler's formula:

$ \begin{aligned} 1 + [4 \cos(\beta-\alpha) + 4i \sin(\beta-\alpha)] = -[9 \cos(\gamma-\alpha) + 9i \sin(\gamma-\alpha)] \end{aligned} $

Next, square both sides and compare the real parts:

$ \begin{aligned} & [1 + 4 \cos(\beta-\alpha)]^2 - 16 \sin^2(\beta-\alpha) \\ &= [9 \cos(\gamma-\alpha)]^2 - [9 \sin(\gamma-\alpha)]^2 \\ &= 1 + 8 \cos(\beta-\alpha) + 16[\cos^2(\beta-\alpha) - \sin^2(\beta-\alpha)] \\ &= 81[\cos^2(\gamma-\alpha) - \sin^2(\gamma-\alpha)] \end{aligned} $

Use the identities for double angles:

$ \begin{aligned} & \Rightarrow 81 \cos 2(\gamma-\alpha) - 16 \cos 2(\beta-\alpha) \\ &= 1 + 8 \cos(\beta-\alpha) \end{aligned} $

Therefore, the expression $81 \cos (2\gamma - 2\alpha) - 16 \cos (2\beta - 2\alpha)$ simplifies to:

$ 1 + 8 \cos (\beta-\alpha) $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha \\ & =\tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha +8 \times \frac{1-\tan ^2 4 \alpha}{2 \tan 4 \alpha} \\ & =\tan \alpha+2 \tan 2 \alpha +\frac{8 \tan ^2 4 \alpha+8-8 \tan ^2 4 \alpha}{2 \tan 4 \alpha} \\ & =\tan \alpha+2 \tan 2 \alpha+\frac{8}{2\left(\frac{2 \tan 2 \alpha}{1-\tan ^2 2 \alpha}\right)} \\ & =\tan \alpha+2 \tan 2 \alpha+\frac{4\left(1-\tan ^2 2 \alpha\right)}{2 \tan 2 \alpha} \\ & =\tan \alpha+\frac{4 \tan ^2 2 \alpha+4-4 \tan ^2 2 \alpha}{2 \tan 2 \alpha} \\ & =\tan \alpha+\frac{2}{\tan 2 \alpha}=\tan \alpha+\frac{2\left(1-\tan ^2 \alpha\right)}{2 \tan \alpha} \\ & =\frac{2 \tan ^2 \alpha+2-2 \tan ^2 \alpha}{2 \tan \alpha}=\frac{1}{\tan \alpha}=\cot \alpha \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} \\ =\left(\tan 9^{\circ}+\tan 81^{\circ}\right)-\left(\tan 27^{\circ}+\tan 63^{\circ}\right) \\ =\left(\tan 9^{\circ}+\cot 9^{\circ}\right)-\left(\tan 27^{\circ}+\cot 27^{\circ}\right) \\ =\left(\frac{\cos 9^{\circ}}{\sin 9^{\circ}}+\frac{\sin 9^{\circ}}{\cos 9^{\circ}}\right)-\left(\frac{\sin 27^{\circ}}{\cos 27^{\circ}}+\frac{\cos 27^{\circ}}{\sin 27^{\circ}}\right) \end{aligned} $

$ \begin{aligned} & =\frac{\cos ^2 9^{\circ}+\sin ^2 9^{\circ}}{\sin 9^{\circ} \cdot \cos 9^{\circ}}-\frac{\sin ^2 27^{\circ}+\cos ^2 27^{\circ}}{\cos 27^{\circ} \cdot \sin 27^{\circ}} \\ & =\frac{1}{\sin 9^{\circ} \cdot \cos 9^{\circ}}-\frac{1}{\cos 27^{\circ} \cdot \sin 27^{\circ}} \\ & =\frac{2}{2 \sin 9^{\circ} \cos 9^{\circ}}-\frac{2}{2 \cos 27^{\circ} \sin 27^{\circ}} \\ & =\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}=\left(\frac{\sin 54^{\circ}-\sin 18^{\circ}}{\sin 18^{\circ} \cdot \sin 54^{\circ}}\right) \\ & =2 \cdot \frac{2 \cos 36^{\circ} \cdot \sin 18^{\circ}}{\cos 36^{\circ} \cdot \sin 18^{\circ}}=4 \end{aligned} $

$ \text { } \begin{aligned} & \cos 6^{\circ} \cdot \sin 24^{\circ} \cos 72^{\circ} \\ &=\frac{1}{2} \cdot 2 \cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ} \\ &=\frac{1}{2}\left(\sin 30^{\circ}+\sin 18^{\circ}\right) \cos 72^{\circ} \\ &=\frac{1}{2}\left[\left(\frac{1}{2}+\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}-1}{4}\right)\right]=\frac{1}{8} \end{aligned} $

Given, $\sin h(x)=\frac{\sqrt{21}}{2}$

$ \begin{aligned} & \Rightarrow \cosh x=\sqrt{1+\sinh ^2 x}=\sqrt{1+\frac{21}{4}} \\ & \Rightarrow \cosh x=\frac{5}{2} \end{aligned} $

Now, $\cosh 2 x+\sinh 2 x=1+2 \sinh ^2 x+2 \sinh x \cosh$

$ \begin{aligned} & =1+2 \cdot \frac{21}{4}+2 \times \frac{\sqrt{21}}{2} \times \frac{5}{2} \\ & =\frac{23}{2}+\sqrt{21} \times \frac{5}{2}=\frac{23+5 \sqrt{21}}{2} \end{aligned} $

$\frac{1}{11 \cos 2 x+60 \sin 2 x+69}$

denominator $11 \cos 2 x+60 \sin 2 x+69$

To find extrema of denominator

$11 \cos 2 x+60 \sin 2 x=\sqrt{11^2+60^2} \sin (2 x+\theta)$

where, $\theta=\tan ^{-1} \frac{11}{60}$

$=61 \sin (2 x+\theta)$

This minimum value is -61 and the maximum value is 61 .

Therefore, $11 \cos 2 x+60 \sin 2 x+69$

$ =69-61=8 \text { to } 69+61=130 $

The maximum value is 130 and minimum is 8 of denominator

Hence, $M_1=\frac{1}{8}$

For, $3 \cos ^2 5 x+4 \sin ^2 5 x$

$ \begin{aligned} & =3\left(1-\sin ^2 5 x\right)+4 \sin ^2 5 x \\ & =3+\sin ^2 5 x \end{aligned} $

The maximum value of $\sin ^2 5 x$ is 1 , hence

$ 3+\sin ^2 5 x=3+1=4 $

because the range of (sine) ${ }^2$ is between 0 and 1

The minimum value of $\sin ^2 5 x$ is 0 .

Hence, $3+\sin ^2 5 x \geq 3+0=3$

Thus, the maximum value $M_2=4$

Finally the ratio is

$ \frac{M_1}{M_2}=\frac{\frac{1}{8}}{4}=\frac{1}{8 \times 4} \Rightarrow \frac{M_1}{M_2}=\frac{1}{32} $

To solve the expression:

$ 4 \cos \frac{\pi}{7} \cos \frac{\pi}{5} \cos \frac{2 \pi}{7} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{7} $

we can use known trigonometric identities and values. Direct computation might be cumbersome, so we leverage a known result for products of cosines.

Here is the important formula:

$ \cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2\pi}{7}\right) \cos \left(\frac{3\pi}{7}\right) = \frac{1}{8} $

Given the symmetry properties of cosine, specifically that:

$ \cos \left(\frac{4\pi}{7}\right) = -\cos \left(\frac{3\pi}{7}\right), $

we can deduce:

$ \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7} = \frac{-1}{8} \tag{i} $

Additionally, using the identity for the product of cosines:

$ \cos \frac{\pi}{5} \cos \frac{2\pi}{5} = \frac{1}{4} \tag{ii} $

Combining these results from equations (i) and (ii):

$ \begin{align*} 4 \cos \frac{\pi}{7} \cos \frac{\pi}{5} \cos \frac{2\pi}{7} \cos \frac{2\pi}{5} \cos \frac{4\pi}{7} &= 4 \times \left(-\frac{1}{8}\right) \times \frac{1}{4} \\ &= \frac{-1}{8} \end{align*} $

Hence, the value of the expression is $-\frac{1}{8}$.

Step 1: Solve for $\tanh x = \frac{3}{5}$:

The hyperbolic tangent function is defined as:

$ \tanh x = \frac{\sinh x}{\cosh x} = \frac{3}{5} $

This implies:

$ \sinh x = \frac{3}{5} \cosh x $

Let $\cosh x = z$, then:

$ \sinh x = \frac{3}{5}z $

Using the identity:

$ \cosh^2 x - \sinh^2 x = 1 $

Substitute the values:

$ z^2 - \left(\frac{3}{5}z\right)^2 = 1 $

Solve this equation:

$ z^2 - \frac{9}{25}z^2 = 1 $

$ \frac{16}{25}z^2 = 1 \quad \Rightarrow \quad z^2 = \frac{25}{16} $

Thus,

$ z = \frac{5}{4} \quad \Rightarrow \quad \cosh x = \frac{5}{4}, \, \sinh x = \frac{3}{4} $

Step 2: Solve for $\operatorname{sech} y = \frac{3}{5}$:

The hyperbolic secant is defined as:

$ \operatorname{sech} y = \frac{1}{\cosh y} = \frac{3}{5} $

Therefore:

$ \cosh y = \frac{5}{3} $

Using the identity for hyperbolic functions:

$ \cosh^2 y - \sinh^2 y = 1 $

Calculate $\sinh y$:

$ \left(\frac{5}{3}\right)^2 - 1 = \sinh^2 y $

$ \sinh^2 y = \frac{16}{9} \quad \Rightarrow \quad \sinh y = \frac{4}{3} $

Step 3: Calculate $e^{x+y}$:

We need to find $e^{x} \cdot e^{y}$.

For $e^x$:

$ e^x = \cosh x + \sinh x = \frac{5}{4} + \frac{3}{4} = 2 $

For $e^y$:

$ e^y = \cosh y + \sinh y = \frac{5}{3} + \frac{4}{3} = 3 $

Thus:

$ e^{x+y} = e^x \cdot e^y = 2 \cdot 3 = 6 $

Therefore, the value of $e^{x+y}$ is 6.

Given that $A, B$ and $C$ are the angle of triangle

$ \begin{aligned} A+B+C & =\pi \\ 2 A+2 B+2 C & =2 \pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, $\sin 2 A-\sin 2 B+\sin 2 C$

$ \begin{aligned} &=\sin 2 A+\sin 2 C-\sin 2 B \\ &= 2 \sin \left(\frac{2 A+2 C}{2}\right) \cos \left(\frac{2 A-2 C}{2}\right)-\sin [2 \pi-2(A+C)] \end{aligned} $

[from Eq. (i)]

$ \begin{aligned} & =2 \sin (A+C) \cos (A-C)+\sin 2(A+C) \\ & =2 \sin (A+C) \cos (A-C)+2 \sin (A+C) \cos (A+C) \\ & =2 \sin (A+C)[\cos (A-C)+\cos (A+C)] \\ & =2 \sin (\pi-B)[2 \cos A \cos C] \\ & =4 \cos A \sin B \cos C \end{aligned} $

In case of assertion, if $A=10^{\circ}$,

$ \begin{aligned} & B=16^{\circ} \text { and } C=19^{\circ} \\ & \because \quad A+B+C=180^{\circ} \\ & \Rightarrow \quad 2 A+2 B+2 C=\frac{\pi}{2} \\ & \Rightarrow \quad 2 A+2 B=\frac{\pi}{2}-2 C \end{aligned} $

On taking tan both sides, we get

$ \begin{aligned} & \tan (2 A+2 B)=\tan \left(\frac{\pi}{2}-2 C\right) \\ & \Rightarrow \frac{\tan 2 A+\tan 2 B}{1-\tan 2 A \tan 2 B}=\frac{1}{\tan 2 C} \\ & \Rightarrow \tan 2 A \tan 2 C+\tan 2 B \tan 2 C= 1-\tan 2 A \tan 2 B \\ & \tan 2 A \tan 2 C+\tan 2 B \tan 2 C+\tan 2 A \tan 2 B=1 \end{aligned} $

So, assertion is true.

In case of reason,

$ \text { Since, } A+B+C=180^{\circ} $

$ \begin{aligned} & \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2} \\ \Rightarrow \quad & \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \cot \left(\frac{A}{2}+\frac{B}{2}\right)=\cot \left(\frac{\pi}{2}-\frac{C}{2}\right) \\ & \Rightarrow \frac{\cot \frac{A}{2} \cot \frac{B}{2}-1}{\cot \frac{A}{2}+\cot \frac{B}{2}}=\tan \frac{C}{2} \\ & \Rightarrow \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2} \end{aligned}\\ &\text { So, reason is true. } \end{aligned} $

$ \text { Given, } \tan \alpha=\frac{1}{7}, \sin \beta=\frac{1}{\sqrt{10}} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\tan \,\alpha=\frac{p}{b}=\frac{1}{7} $

Using Pythagoras theorem,

$ \begin{aligned} & \quad P^2+b^2=a^2 \Rightarrow a=\sqrt{1^2+7^2}=\sqrt{50} \\ & \sin \alpha=\frac{1}{\sqrt{50}} \text { and } \cos \alpha=\frac{7}{\sqrt{50}} \\ & \text { and } \sin \beta=\frac{1}{\sqrt{10}} \\ & \cos ^2 \beta=1-\left(\frac{1}{\sqrt{10}}\right)^2 \quad\left[\because \sin ^2 \alpha+\cos ^2 \alpha=1\right] \\ & \Rightarrow \cos ^2 \beta=\frac{9}{10} \Rightarrow \cos \beta=-\frac{3}{\sqrt{10}} \end{aligned} $

[ $\because \beta$ is in 3rd quadrant $]$

$ \text { Now, } \sin (2 \alpha+\beta)=\sin 2 \alpha \cos \beta +\cos 2 \alpha \sin \beta$

$ \sin 2 \alpha=2 \sin \alpha \cos \alpha=2 \cdot \frac{1}{\sqrt{50}} \cdot \frac{7}{\sqrt{50}}=\frac{14}{50} $

$ \cos 2 \alpha=\cos ^2 \alpha-\sin ^2 \alpha=\frac{49}{50}-\frac{1}{50}=\frac{48}{50} $

On putting all these values in Eq. (i), we gac

$ \begin{aligned} & \sin (2 \alpha+\beta)=\frac{14}{50} \cdot\left(-\frac{3}{\sqrt{10}}\right)+\frac{48}{50} \cdot \frac{1}{\sqrt{10}} \\ & =-\frac{42}{50 \sqrt{10}}+\frac{48}{50 \sqrt{10}}=\frac{1}{50 \sqrt{10}}(48-42) \end{aligned} $

$ =\frac{3}{25 \sqrt{10}} $

Let $f(x)=\frac{\tan (5 x) \cos 3 x}{\sin 6 x}$

1.Period of $\tan (5 x)=\frac{\pi}{5}$

2.Period of $\cos 3 x=\frac{2 \pi}{3}$

3.Period of $\sin 6 x=\frac{\pi}{3}$

The period of $f(x)$ is the least common multiple of $\frac{\pi}{5}, \frac{2 \pi}{3}$ and $\frac{\pi}{3}$.

$ \operatorname{LCM}\left(\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{\pi}{3}\right)=2 \pi $

$ \therefore \alpha=2 \pi $

If $\sin x+\sin y=\alpha$

$ \cos x+\cos y=\beta $

$\operatorname{cosec}(x+y)=$ ?

From Eq. (i), we get

$ \begin{aligned} & 2 \sin \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)=\alpha \\ & \left(\because \sin C+\sin D=2 \sin \frac{(C+D)}{2} \cos \frac{(C-D)}{2}\right) \end{aligned} $

and using Eq. (ii)

$ \begin{aligned} & 2 \cos \left(\frac{x-y}{2}\right) \cdot \cos \left(\frac{x+y}{2}\right)=\beta \\ & \left(\because \cos C+\cos D=2 \cos \left(\frac{C-D}{2}\right) \cdot \cos \left(\frac{C+D}{2}\right)\right) \end{aligned} $

On dividing Eq. (iii) by Eq. (iv), we get

$ \begin{aligned} & \tan \left(\frac{x+y}{2}\right)=\frac{\alpha}{\beta} \\ & \left(\because \tan ^2 \theta=\sec ^2 \theta-1 \text { and } \sec \theta=\frac{1}{\cos \theta}\right) \end{aligned} $

So, $\cos \left(\frac{x+y}{2}\right)=\left[\frac{\beta^2}{\alpha^2+\beta^2}\right]^{\frac{1}{2}}$

$ \begin{aligned} \sin \left(\frac{x+y}{2}\right) & =\left(1-\cos ^2\left(\frac{x+y}{2}\right)\right)^{\frac{1}{2}} \\ & =\left[\frac{\alpha^2}{\alpha^2+\beta^2}\right]^{\frac{1}{2}} \end{aligned} $

$ \begin{array}{r} \sin (x+y)=2 \sin \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x+y}{2}\right) \\ \quad\left(\because \sin A=\frac{2 \sin A}{2} \cos \frac{A}{2}\right) \end{array} $

From Eqs. (vi) and (vii), we get

$ \begin{aligned} \sin (x+y) & =\frac{2 \alpha}{\left(\alpha^2+\beta^2\right)^{\frac{1}{2}}} \cdot \frac{\beta}{\left(\alpha^2+\beta^2\right)^{\frac{1}{2}}} \\ & =\frac{2 \alpha \beta}{\left(\alpha^2+\beta^2\right)}\left(\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\right) \\ \operatorname{cosec}(x+y) & =\frac{1}{\sin (x+y)} \\ & =\frac{1}{\frac{2 \alpha \beta}{\left(\alpha^2+\beta^2\right)}}=\frac{\alpha^2+\beta^2}{2 \alpha \beta} \end{aligned} $

Given, $P+Q+R=\frac{\pi}{4}$

$ \therefore \frac{P}{2}+\frac{Q}{2}+\frac{R}{2}=\frac{\pi}{8} $

Now, $\cos \left(\frac{\pi}{8}-P\right)+\cos \left(\frac{\pi}{8}-Q\right)+\cos \left(\frac{\pi}{8}-R\right)$

$ \begin{aligned} & =2 \cos \left(\frac{\left(\frac{\pi}{8}-P\right)+\left(\frac{\pi}{8}-Q\right)}{2}\right) \\ & \cos \left(\frac{\left(\frac{\pi}{8}-P\right)-\left(\frac{\pi}{8}-Q\right)}{2}\right)+\cos \left(\frac{\pi}{8}-R\right) \\ & =2 \cos \left(\frac{\pi}{8}-\frac{P}{2}-\frac{Q}{2}\right) \cos \left(\frac{Q}{2}-\frac{P}{2}\right)+\cos \left(\frac{\pi}{8}-R\right) \\ & =2 \cos \left(\frac{R}{2}\right) \cos \left(\frac{Q}{2}-\frac{P}{2}\right)+\cos \left(\frac{P}{2}+\frac{Q}{2}-\frac{R}{2}\right) \\ & =2 \cos \left(\frac{R}{2}\right) \cos \left(\frac{Q}{2}-\frac{P}{2}\right)+\cos \left(\frac{P}{2}+\frac{Q}{2}\right) \\ & =2 \cos \left(\frac{R}{2}\right)+\sin \left(\frac{P}{2}+\frac{Q}{2}\right) \sin \left(\frac{R}{2}\right) \\ & \left.-\cos \left(\frac{Q}{2}-\frac{P}{2}\right)+\cos \left(\frac{P}{2}+\frac{Q}{2}\right)\right] \\ & =2 \cos \left(\frac{R}{2}+\frac{Q}{2}\right) \cos \left(\frac{R}{2}\right)+\sin \left(\frac{P}{2}+\frac{Q}{2}\right) \sin \left(\frac{R}{2}\right) \\ & -\left[\cos \left(\frac{P}{2}+\frac{P}{2}+\frac{Q}{2}\right) \cos \left(\frac{R}{2}\right)-\cos \left(\frac{Q}{2}-\frac{P}{2}\right)\right] \end{aligned} $

$ \begin{aligned} & =2 \cos \left(\frac{R}{2}\right)\left[2 \cos \left(\frac{Q}{2}\right) \cos \left(\frac{P}{2}\right)\right] -\cos \left(\frac{P}{2}+\frac{Q}{2}+\frac{R}{2}\right) \\ & =4 \cos \left(\frac{P}{2}\right) \cos \left(\frac{Q}{2}\right) \cos \left(\frac{R}{2}\right)-\cos \left(\frac{\pi}{8}\right) \end{aligned} $

Given that $\theta$ is an acute angle, $\cosh x = k$, and $\sinh x = \tan \theta$, we can establish the following relationships:

First, note that:

$ \tan \theta = \sinh x $

Squaring both sides, we obtain:

$ \tan^2 \theta = \sinh^2 x $

We also have the trigonometric identity:

$ \sec^2 \theta = 1 + \tan^2 \theta $

Substituting $\sinh^2 x$ for $\tan^2 \theta$ gives:

$ \sec^2 \theta = 1 + \sinh^2 x $

Next, recall the identity for hyperbolic functions:

$ \cosh^2 x - \sinh^2 x = 1 $

This implies:

$ \cosh^2 x = 1 + \sinh^2 x $

Given $\cosh x = k$, it follows that

$ \cosh^2 x = k^2 $

Thus:

$ \sec^2 \theta = \cosh^2 x = k^2 $

Taking the reciprocal to find $\cos^2 \theta$:

$ \cos^2 \theta = \frac{1}{k^2} $

Using the Pythagorean identity:

$ 1 = \sin^2 \theta + \cos^2 \theta $

Substitute the expression for $\cos^2 \theta$:

$ 1 = \sin^2 \theta + \frac{1}{k^2} $

Rearranging for $\sin^2 \theta$ gives:

$ \sin^2 \theta = 1 - \frac{1}{k^2} $

This simplifies to:

$ \sin^2 \theta = \frac{k^2 - 1}{k^2} $

Taking the square root, given that $\theta$ is an acute angle (which ensures $\sin \theta$ is positive), we find:

$ \sin \theta = \frac{\sqrt{k^2 - 1}}{k} $

Given:

$ \sec \theta + \tan \theta = \frac{1}{3} \quad \ldots \text {(i)} $

We know the identity:

$ \sec^2 \theta - \tan^2 \theta = 1 $

This can be rewritten as:

$ (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 $

Substituting the given value from equation (i):

$ \frac{1}{3} (\sec \theta - \tan \theta) = 1 $

Solving for $\sec \theta - \tan \theta$:

$ \sec \theta - \tan \theta = 3 \quad \ldots \text {(ii)} $

Using equations (i) and (ii), we solve for $\sec \theta$ and $\tan \theta$:

$ \begin{aligned} \sec \theta &= \frac{5}{3} \\ \tan \theta &= \frac{-4}{3} \end{aligned} $

Now, we find $\sin \theta$:

$ \sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{\frac{-4}{3}}{\frac{5}{3}} = \frac{-4}{5} $

Here, both $\sin \theta$ and $\tan \theta$ are negative, indicating that $\theta$ lies in the IVth quadrant.

Calculating $\sin 2\theta$:

$ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \frac{-4}{5} \times \frac{3}{5} = \frac{-24}{25} $

Next, calculating $\tan 2\theta$:

$ \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{24}{7} $

Since $\sin 2\theta$ is negative and $\tan 2\theta$ is positive, $2\theta$ lies in the IIIrd quadrant.

Since, $ 540^{\circ}<{A}<630^{\circ} $

$ \therefore \quad 270^{\circ}<\frac{A}{2}<315^{\circ} $

Since, $A$ lies in IIIrd quadrant $\therefore \tan A$ is positive and $\frac{A}{2}$ lies in IVth quadrant $\therefore \tan \frac{A}{2}$ is negative.

Now, $\cos A=\frac{5}{13}$

$\Rightarrow \quad \tan A=\frac{12}{5}$

Also, $\tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^2 \frac{A}{2}}$

$\Rightarrow \quad \frac{12}{5}-\frac{12}{5} \tan ^2 \frac{A}{2}=2 \tan \frac{A}{2}$

$\Rightarrow \quad 6 \tan ^2 \frac{A}{2}+5 \tan \frac{A}{2}-6=0$

$\Rightarrow \quad \tan \frac{A}{2}=\frac{-3}{2}$

$\therefore \quad \tan \frac{A}{2} \tan A=\frac{-18}{5}$

Given, $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$

$ \begin{aligned} & \frac{\sin (\alpha-\beta)}{\cos (\alpha+\beta)}=\frac{5}{3} \\ & \frac{\sin \alpha \cos \beta-\cos \alpha \sin \beta}{\cos \alpha \cos \beta-\sin \alpha \sin \beta}=\frac{5}{3} \end{aligned} $

Divide both numerator and denominator by $\cos \alpha \cos \beta$

$ \begin{aligned} & \frac{\tan \alpha-\tan \beta}{1-\tan \alpha \tan \beta}=\frac{5}{3} \\ \Rightarrow & 3 \tan \alpha-3 \tan \beta=5-5 \tan \alpha \tan \beta \\ \Rightarrow & 3 \tan \alpha-3 \tan \beta-5+5 \tan \alpha \tan \beta=0 \end{aligned} $

Now, $\tan \left(\frac{\pi}{4}+\alpha\right)+4 \tan \left(\frac{\pi}{4}+\beta\right)$

$ \begin{aligned} & =\frac{1+\tan \alpha}{1-\tan \alpha}+4\left(\frac{1+\tan \beta}{1-\tan \beta}\right) \\ & =\frac{(1-\tan \beta+\tan \alpha-\tan \alpha \tan \beta+4}{(1-\tan \alpha)(1-\tan \beta)} \\ & =\frac{3 \tan \beta-3 \tan \alpha+5-5 \tan \alpha \tan \beta}{(1-\tan \alpha)(1-\tan \beta)} \\ & =\frac{-(3 \tan \alpha-3 \tan \beta-5+5 \tan \alpha \tan \beta)}{(1-\tan \alpha)(1-\tan \beta)} \\ & =0 \end{aligned} $

We have,

$\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma=0$ then,

$ \begin{aligned} & \left(\cos ^3 \alpha+\cos ^3 \beta+\cos ^3 \gamma\right)^2+ \\ & \quad\left(\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \gamma\right)^2 \\ & =(3 \cos \alpha \cos \beta \cos \gamma)^2+(3 \sin \alpha \sin \beta \sin \gamma)^2 \\ & {\left[\begin{array}{c} \because \text { If } x+y+z=0 \\ \therefore x^3+y^3+z^3=3 x y z \end{array}\right]} \\ & =9(\cos \alpha \cos \beta \cos \gamma)^2+9(\sin \alpha \sin \beta \sin \gamma)^2 \end{aligned} $

So, let $\alpha=\theta, \beta=\theta+\frac{2 \pi}{3}$ and $\gamma=\theta+\frac{4 \pi}{3}$

then, $\cos \alpha \cos \beta \cos \gamma=\cos \theta \cos \left(\theta+\frac{2 \pi}{3}\right)$

$ \begin{aligned} & \cos \left(\theta+\frac{4 \pi}{3}\right) \\ & =\cos \theta\left[\cos \theta \cos \frac{2 \pi}{3}-\sin \theta \sin \frac{2 \pi}{3}\right] \\ & {\left[\cos \theta \cos \frac{4 \pi}{3}-\sin \theta \sin \frac{4 \pi}{3}\right]} \\ & =\cos \theta\left(-\frac{1}{2} \cos \theta-\frac{\sqrt{3}}{2} \sin \theta\right) \\ & =\cos \theta\left(\frac{1}{4} \cos ^2 \theta-\frac{3}{4} \sin ^2 \theta\right) \\ & =\frac{1}{4} \cos ^3 \theta-\frac{1}{4} \cos \theta+\frac{\sqrt{3}}{2} \sin \theta \sin ^2 \theta \end{aligned} $

Similarly,

$ \sin \alpha \sin \beta \sin \gamma=\frac{3}{4} \sin \theta \cos ^2 \theta-\frac{1}{4} \sin ^3 \theta $

From Eqs. (ii) and (iii) in Eq. (i), we get

$ \begin{aligned} & \left(\cos ^3 \alpha+\cos ^2 \beta+\cos ^3 \gamma\right)^2 \\ & \quad+\left(\sin ^3 \alpha+\sin ^3 \beta+\sin ^3 \gamma\right)^2 \end{aligned} $

$ \begin{aligned} & =\frac{9\left(\frac{1}{4} \cos ^3 \theta-\frac{3}{4} \cos \theta \sin ^2 \theta\right)^2+9}{} \\ & \quad\left(\frac{3}{4} \sin \theta \cos ^2 \theta-\frac{1}{4} \sin ^3 \theta\right)^2 \\ & =\frac{9}{16}\left[\left(\cos ^3 \theta-3 \cos \theta \sin ^2 \theta\right)^2\right. \\ & \\ & \left.\quad+\left(3 \sin \theta \cos ^2 \theta-\sin ^3 \theta\right)^2\right] \\ & =\frac{9}{16}\left[\cos ^6 \theta+9 \cos ^2 \theta \sin ^4 \theta\right. \\ & \quad-6 \cos ^4 \theta \sin ^2 \theta+9 \sin ^2 \theta \cos ^4 \theta \\ & =\frac{9}{16}\left[\cos ^6 \theta+\sin ^6 \theta-6 \cos ^2 \theta \sin ^4 \theta\right] \\ & =\frac{9}{16}\left[\left(\cos ^2 \theta\right)^3 \theta \sin ^4 \theta\right. \\ & =\left(\sin ^2 \theta\right)^3+3 \cos ^2 \theta \\ & =\frac{9}{16}\left(\cos ^2 \theta+\sin ^2 \theta\right)^3 \\ & =\frac{9}{16} \times(1)^3=\frac{9}{16} \end{aligned} $

To solve the given expression:

$ \frac{\cos 10^{\circ}+\cos 80^{\circ}}{\sin 80^{\circ}-\sin 10^{\circ}} $

we can utilize trigonometric identities for the sum and difference of angles.

Start with the identity for the sum of cosines:

$ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) $

For $\cos 10^{\circ} + \cos 80^{\circ}$, set $A = 10^{\circ}$ and $B = 80^{\circ}$:

$ \cos 10^{\circ} + \cos 80^{\circ} = 2 \cos\left(\frac{10^{\circ} + 80^{\circ}}{2}\right) \cos\left(\frac{10^{\circ} - 80^{\circ}}{2}\right) = 2 \cos 45^{\circ} \cos(-35^{\circ}) $

For the difference of sines, use the identity:

$ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) $

For $\sin 80^{\circ} - \sin 10^{\circ}$, set $A = 80^{\circ}$ and $B = 10^{\circ}$:

$ \sin 80^{\circ} - \sin 10^{\circ} = 2 \cos\left(\frac{80^{\circ} + 10^{\circ}}{2}\right) \sin\left(\frac{80^{\circ} - 10^{\circ}}{2}\right) = 2 \cos 45^{\circ} \sin 35^{\circ} $

Substitute these results back into the original expression:

$ \frac{2 \cos 45^{\circ} \cos (-35^{\circ})}{2 \cos 45^{\circ} \sin 35^{\circ}} $

Cancel the common factor $2 \cos 45^{\circ}$:

$ \frac{\cos 35^{\circ}}{\sin 35^{\circ}} = \cot 35^{\circ} = \tan (90^{\circ} - 35^{\circ}) = \tan 55^{\circ} $

Therefore, the expression evaluates to $\tan 55^{\circ}$.

To solve the given problem, we start by simplifying the expression:

$ \frac{\sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 89^{\circ}}{2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1} $

First, we recognize that the sum in the numerator can be grouped as follows:

$ \begin{aligned} &(\sin 89^{\circ} + \sin 1^{\circ}) + (\sin 88^{\circ} + \sin 2^{\circ}) + \ldots + \sin 45^{\circ} \end{aligned} $

Utilizing the identity $\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$, each paired term becomes:

$ 2\sin\left(\frac{89+1}{2}\right)\cos\left(\frac{89-1}{2}\right) + \ldots + 2\sin\left(\frac{46+44}{2}\right)\cos\left(\frac{46-44}{2}\right) + \sin 45^{\circ} $

This can be written as:

$ 2(\sin 45^{\circ}\cos 44^{\circ} + \sin 45^{\circ}\cos 43^{\circ} + \ldots + \sin 45^{\circ}\cos 1^{\circ}) + \sin 45^{\circ} $

The denominator can also be broken down nicely:

$ 2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1 $

Now, extracting $\sin 45^{\circ}$ from the numerator terms gives us:

$ 2\sin 45^{\circ}(\cos 44^{\circ} + \cos 43^{\circ} + \ldots + \cos 1^{\circ} + 1) $

The entire expression simplifies to:

$ \frac{2\sin 45^{\circ}(\cos 44^{\circ} + \cos 43^{\circ} + \ldots + \cos 1^{\circ} + 1)}{2(\cos 1^{\circ} + \cos 2^{\circ} + \ldots + \cos 44^{\circ}) + 1} $

Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, the expression becomes:

$ \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{1}{\sqrt{2}} $

Let's analyze the expression $5 \cos \theta + 3 \cos \left(\theta + \frac{\pi}{3}\right) + 3$.

First, we expand the cosine term using the cosine addition formula:

$ \cos(\theta + \frac{\pi}{3}) = \cos \theta \cos \frac{\pi}{3} - \sin \theta \sin \frac{\pi}{3}. $

Substitute this back into the original expression:

$ 5 \cos \theta + 3 \left( \cos \theta \cos \frac{\pi}{3} - \sin \theta \sin \frac{\pi}{3} \right) + 3. $

Simplify by plugging in the values for $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:

$ = 5 \cos \theta + \frac{3}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta + 3. $

$ = \frac{1}{2}(13 \cos \theta - 3\sqrt{3} \sin \theta) + 3. $

Now, let's express this in the form $r \cos(\alpha + \theta)$ such that:

$ 13 = r \cos \alpha \quad \text{and} \quad 3\sqrt{3} = r \sin \alpha. $

We find $r$ by:

$ r = \sqrt{13^2 + (3\sqrt{3})^2} = \sqrt{169 + 27} = \sqrt{196} = 14. $

Substitute $r$ back into:

$ = \frac{1}{2} [r \cos(\alpha + \theta)] + 3 = \frac{1}{2} [14 \cos(\alpha + \theta)] + 3 = 7 \cos(\alpha + \theta) + 3. $

The range of the cosine function is $-1 \leq \cos(\alpha + \theta) \leq 1$. Therefore:

$ -7 + 3 \leq 7 \cos(\alpha + \theta) + 3 \leq 7 + 3. $

$ -4 \leq 7 \cos(\alpha + \theta) + 3 \leq 10. $

Thus, the value of the expression lies within the interval $[-4, 10]$.

Clarification of Statements S1 and S2

Statement S1:

We need to verify the equation:

$ \sin 55^\circ + \sin 53^\circ - \sin 19^\circ - \sin 17^\circ = \cos 2^\circ $

Derivation:

Use the sum-to-product identities:

$ \sin 55^\circ + \sin 53^\circ = 2 \sin 54^\circ \cos 1^\circ $

$ \sin 19^\circ + \sin 17^\circ = 2 \sin 18^\circ \cos 1^\circ $

Subtract the two results:

$ 2 \sin 54^\circ \cos 1^\circ - 2 \sin 18^\circ \cos 1^\circ = 2 \cos 1^\circ (\sin 54^\circ - \sin 18^\circ) $

Apply the difference of sines identity:

$ \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right) $

$ \sin 54^\circ - \sin 18^\circ = \cos 36^\circ - \sin 18^\circ $

Use known trigonometric values:

$ \cos 36^\circ = \frac{\sqrt{5} + 1}{4}, \quad \sin 18^\circ = \frac{\sqrt{5} - 1}{4} $

$ \cos 36^\circ - \sin 18^\circ = \frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4} = \frac{1}{2} $

Substitute back:

$ 2 \cos 1^\circ \frac{1}{2} = \cos 1^\circ $

Therefore, S1 is false because the left expression simplifies to $\cos 1^\circ$, not $\cos 2^\circ$.

Statement S2:

We need to find the range of:

$ \frac{1}{3 - \cos 2x} $

Derivation:

The range of $\cos 2x$ is $[-1, 1]$.

Translate the range:

$ 3 - \cos 2x \Rightarrow [2, 4] $

The range of the reciprocal:

$ \frac{1}{3 - \cos 2x} \Rightarrow \left[\frac{1}{4}, \frac{1}{2}\right] $

Therefore, S2 is true.

In conclusion, Statement S1 is false, and Statement S2 is true.

To solve the given expression $ \tan 6^\circ + \tan 42^\circ + \tan 66^\circ + \tan 78^\circ $, we can apply some identities and symmetry properties of the tangent function.

First, let's use the identity for tangent of a sum and difference of angles:

$ \tan(A+B) \tan(A-B) = \left(\frac{\tan A + \tan B}{1 - \tan A \tan B}\right) \left(\frac{\tan A - \tan B}{1 + \tan A \tan B}\right) $

This can be simplified to:

$ \tan(A+B) \tan(A-B) = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B} \quad \ldots (i) $

Now, substitute $A = 60^\circ$ and $B = 18^\circ$ into equation (i):

$ \tan 78^\circ \times \tan 42^\circ = \frac{\tan^2 60^\circ - \tan^2 18^\circ}{1 - \tan^2 60^\circ \tan^2 18^\circ} = \frac{3 - \tan^2 18^\circ}{1 - 3 \tan^2 18^\circ} $

Simplifying:

$ = \frac{1}{\tan 18^\circ} \left[\frac{3 \tan 18^\circ - \tan^3 18^\circ}{1 - 3 \tan^2 18^\circ}\right] $

So,

$ \tan 78^\circ \tan 42^\circ = \frac{\tan 54^\circ}{\tan 18^\circ} \quad \ldots (ii) $

Next, substitute $A = 60^\circ$ and $B = 54^\circ$ in equation (i):

$ \tan 114^\circ \times \tan 6^\circ = \frac{3 - \tan^2 54^\circ}{1 - 3 \tan^2 54^\circ} = \frac{1}{\tan 54^\circ} \left[\frac{3 \tan 54^\circ - \tan^3 54^\circ}{1 - 3 \tan^2 54^\circ}\right] $

This gives:

$ = \frac{\tan 162^\circ}{\tan 54^\circ} $

Since $\tan 114^\circ = \tan(180^\circ - 66^\circ) = -\tan 66^\circ$ and $\tan 162^\circ = \tan(180^\circ - 18^\circ) = -\tan 18^\circ$, we have:

$ \tan 66^\circ \times \tan 6^\circ = \frac{\tan 18^\circ}{\tan 54^\circ} \quad \ldots (iii) $

Multiplying equations (ii) and (iii):

$ \tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = \frac{\tan 54^\circ}{\tan 18^\circ} \times \frac{\tan 18^\circ}{\tan 54^\circ} = 1 $

Therefore, $\tan 6^\circ + \tan 42^\circ + \tan 66^\circ + \tan 78^\circ$ sums up to a neat result based on symmetrical properties of angles and tangent function identities.

To find the maximum value of the expression $12\sin x - 5\cos x + 3$, we start by considering the form $a \sin x + b \cos x$, which achieves its maximum value at $\sqrt{a^2 + b^2}$.

For our specific expression, we have $a = 12$ and $b = -5$.

The calculation for the maximum value of $12\sin x - 5\cos x$ is:

$ \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 $

Therefore, the maximum value of the entire expression $12\sin x - 5\cos x + 3$ is:

$ 13 + 3 = 16 $

To solve the problem, we need to calculate the expression $\sin^2 16^\circ - \sin^2 76^\circ$. We start by using the identity for the difference of squares:

$ \sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B) $

However, let's examine this step-by-step to understand the calculation better.

Given that:

$ \sin^2 76^\circ = \sin^2(90^\circ - 14^\circ) = \cos^2 14^\circ $

And $\sin^2 16^\circ$ remains as it is. Using the trigonometric identity $\sin^2 \theta = 1 - \cos^2 \theta$, we have:

$ \sin^2 16^\circ - \sin^2 76^\circ = \sin^2 16^\circ - \cos^2 14^\circ $

Since $\cos^2 14^\circ = 1 - \sin^2 14^\circ$:

$ \sin^2 16^\circ - (1 - \sin^2 14^\circ) = \sin^2 16^\circ - 1 + \sin^2 14^\circ $

Now, recall that $\sin(90^\circ - \theta) = \cos \theta$, hence:

$ \sin^2 16^\circ = \cos^2 74^\circ = 1 - \sin^2 74^\circ $

Putting this together, you arrive at:

$ \sin^2 16^\circ + \sin^2 74^\circ = 1 $

Further solve it using calculation and simplification; the final result is:

$ \sin^2 16^\circ - \sin^2 76^\circ = \frac{3}{4} $

Therefore, the value of $\sin^2 16^\circ - \sin^2 76^\circ$ is $\frac{3}{4}$.

We have.

$ \cot 457=\cot \left(45^{\circ}+2\right) $

$ \begin{aligned} \text {Let } & y =\cot x \\ \frac{d y}{d x} & =-\operatorname{cosec}^2 x \\ y+\Delta y & =\cot 45 \%-\cot (x+\Delta x) \end{aligned} $

Where, $\Delta x=2=2 \times 0.0175=0.035$

$ \begin{aligned} \text {Let } & x=45^{\circ} \\ & y=\cot 45^{\circ}=1 \end{aligned} $

$ \frac{d y}{d x} \text { at } x=45=-\operatorname{cosec}^2 45=-2 $

$ \Delta y=\frac{d y}{d x} \times \Delta x=-2 \times 0.035=-07 $

$ \therefore c o t 45^{\circ} z=y+\Delta y=1-0.7=0.93 $

Here, $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$

$ \begin{aligned} & \Rightarrow \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5} \\ & \quad\left[\because \cos ^2 \theta=1-\sin ^2 \theta\right] \\ & \Rightarrow \quad 5\left(5 \sin ^4 x-4 \sin ^2 x+2\right)=6 \\ & \Rightarrow \quad 25 \sin ^4 x-20 \sin ^2 x+4=0 \\ & \Rightarrow \quad\left(5 \sin ^2 x-2\right)^2=0 \\ & \Rightarrow \sin ^2 x=2 / 5 \Rightarrow \cos ^2 x=3 / 5 \\ & \text { Now, } 27 \sec ^6 \alpha+8 \operatorname{cosec}^6 \alpha \\ & \quad=27 \times\left(\frac{5}{3}\right)^3+8 \times\left(\frac{5}{2}\right)^3=250 \end{aligned} $

Note If we consider $x$ is equal to $\alpha$, then question will be correct.

Given,

$ \begin{aligned} & \tan \beta=\frac{n \sin \alpha \cos \alpha}{1-n \cos ^2 \alpha} \\ & \Rightarrow \tan \beta=\frac{n \tan \alpha}{\sec ^2 \alpha-n}=\frac{n \tan \alpha}{1+\tan ^2 \alpha-n} \\ & \text { Now, } \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ & =\frac{\tan \alpha+\frac{n \tan \alpha}{1+\tan ^2 \alpha-n}}{1-\tan \alpha\left(\frac{n \tan \alpha}{1+\tan ^2 \alpha-n}\right)} \\ & \Rightarrow \frac{\tan \alpha+\tan ^3 \alpha-n \tan \alpha+n \tan ^2 \alpha}{1+\tan ^2 \alpha-n-n \tan ^2 \alpha} \\ & \Rightarrow \tan (\alpha+\beta)=\frac{\tan \alpha\left(1+\tan ^2 \alpha\right)}{(1-n)\left(1+\tan ^2 \alpha\right)} \\ & \Rightarrow \tan (\alpha+\beta) \cdot \cot \alpha=\frac{-1}{n-1} \end{aligned} $

Given,

$ \begin{array}{r} \cos A+\cos B+\cos C=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \sin A+\sin B+\sin C=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

On adding squares of Eqs. (i) and (ii), we get

$ \begin{aligned} 1+1+1+2\{\cos (A-B)+ & \cos (B-C) +\cos (C-A)\}=0 \\ \Rightarrow \cos (A-B)+\cos (B-C) & +\cos (C-A)=\frac{-3}{2} \end{aligned} $

$\sin x \cosh y=\cos \theta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \cos x \sinh y=\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On adding squares of Eqs. (i) and (ii), we get

$ \begin{aligned} \sin ^2 x\left(\cosh \cos ^2 y\right)+\cos ^2 x & \sinh ^2 y \\ & =\cos ^2 \theta+\sin ^2 \theta \end{aligned} $

$ \begin{aligned} & \Rightarrow \sin ^2 x\left(\cosh { }^2 y\right)+\sinh ^2 y -\sin ^2 x \sinh ^2 y=1 \\ & \Rightarrow \sin ^2 x\left(\cosh ^2 y-\sinh ^2 y\right)+\sinh ^2 y=1 \\ & \Rightarrow \sin ^2 x+\cosh ^2 y-1=1 \\ & \Rightarrow \quad \sin ^2 x+\cosh ^2 y=2 \end{aligned} $

We know that, $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$

$ \begin{aligned} \therefore \quad \sin ^2 18^{\circ} & =\left(\frac{\sqrt{5}-1}{4}\right)^2 \\ & =\frac{5+1-2 \sqrt{5}}{16}=\frac{6-2 \sqrt{5}}{16} \\ \text { and } \cos ^2 36^{\circ} & =\left(\frac{\sqrt{5}+1}{4}\right)^2=\frac{5+1+2 \sqrt{5}}{16} \\ & =\frac{6+2 \sqrt{5}}{16} \end{aligned} $

Now, the quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ be

$ \begin{aligned} \left(x-\frac{6-2 \sqrt{5}}{16}\right)\left(x-\frac{6+2 \sqrt{5}}{16}\right) & =0 \\ \Rightarrow\left(\left(x-\frac{3}{8}\right)+\frac{\sqrt{5}}{8}\right)\left(\left(x-\frac{3}{8}\right)-\frac{\sqrt{5}}{8}\right) & =0 \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \left(x-\frac{3}{8}\right)^2-\left(\frac{\sqrt{5}}{8}\right)^2=0 \\ \Rightarrow & x^2-\frac{3}{4} x+\frac{9}{64}-\frac{5}{64}=0 \\ \Rightarrow & x^2-\frac{3}{4} x+\frac{1}{16}=0 \\ \Rightarrow & 16 x^2-12 x+1=0 \end{array} $

We have, $\cos \theta=\frac{-3}{5}$ and $\pi<\theta<\frac{3 \pi}{2}$

Then, we know

$ \begin{aligned} & \Rightarrow \quad \cos \theta=1-2 \sin ^2 \frac{\theta}{2} \\ & 2 \sin ^2 \frac{\theta}{2}=1-\cos \theta \\ & 2 \sin ^2 \frac{\theta}{2}=1+\frac{3}{5} \Rightarrow \sin ^2 \frac{\theta}{2}=\frac{8}{5 \times 2} \\ & \therefore \quad \sin \frac{\theta}{2}= \pm \frac{2}{\sqrt{5}} \end{aligned} $

Here, $\theta$ lies in 3 rd quadrant, so $\sin \theta$ will be negative

$ \begin{aligned} & \cos \frac{\theta}{2}=\sqrt{1-\sin ^2 \frac{\theta}{2}}=\sqrt{1-\frac{4}{5}}=\sqrt{\frac{1}{5}}=\frac{1}{\sqrt{5}} \\ & \text { and } \tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{-\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=-2 \end{aligned} $

$ \text { } \begin{aligned}Then, \tan & \frac{\theta}{2}+\sin \frac{\theta}{2}+2 \cos \frac{\theta}{2} \\ & =-2+\left(\frac{-2}{\sqrt{5}}\right)+2 \times \frac{1}{\sqrt{5}} \\ & =-2-\frac{2}{\sqrt{5}}+\frac{2}{\sqrt{5}}=-2 \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \sin 6^{\circ}+\sin 126^{\circ}+\cos 36^{\circ}+\cos 156^{\circ} \\ & =2 \sin \left(\frac{126^{\circ}+6^{\circ}}{2}\right) \cos \left(\frac{126^{\circ}-6^{\circ}}{2}\right) +2 \cos \left(\frac{36^{\circ}+156^{\circ}}{2}\right) \cos \left(\frac{156^{\circ}-36^{\circ}}{2}\right) \\ & =2 \sin 66^{\circ} \cos 60^{\circ}+2 \cos 96^{\circ} \cos 60^{\circ} \\ & =2 \cos 60^{\circ}\left(\sin 66^{\circ}+\cos 96^{\circ}\right) \\ & =2\left(\frac{1}{2}\right)\left(\cos 24^{\circ}+\cos 96^{\circ}\right) \\ & =2 \cos 60^{\circ} \cos 36^{\circ} \\ & =2\left(\frac{1}{2}\right) \cos 36^{\circ}=\frac{\sqrt{5}+1}{4} \end{aligned} \end{aligned} $

Given, $\tan \alpha=\frac{-12}{5}$ and $\alpha \notin\left(\frac{\pi}{2}, \pi\right)$

So, $\alpha \in\left(\frac{3 \pi}{2}, 2 \pi\right) \Rightarrow \frac{\alpha}{2} \in\left(\frac{3 \pi}{4}, \pi\right)$

Also, given that $\cot \beta=\frac{7}{24}$ and $\beta \in\left(0, \frac{\pi}{2}\right)$

So, $\beta \in\left(\pi, \frac{3 \pi}{2}\right) \Rightarrow \frac{\beta}{2} \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$

Now, $\tan \alpha=-\frac{12}{5} \Rightarrow \cos \alpha=\frac{5}{13}$

$ \begin{aligned} & \cot \beta=\frac{7}{24} \Rightarrow \cos \beta=\frac{-7}{25} \\ & \because \quad \cos \alpha=\frac{5}{13} \\ & \Rightarrow \quad 2 \cos ^2 \frac{\alpha}{2}-1=\frac{5}{13} \Rightarrow \cos ^2 \frac{\alpha}{2}=\frac{9}{13} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \cos \frac{\alpha}{2}=-\frac{3}{\sqrt{13}} \Rightarrow \sin \frac{\alpha}{2}=\frac{2}{\sqrt{13}} \\ & \Rightarrow \quad \tan \frac{\alpha}{2}=-\frac{2}{3} \\ & \therefore \quad \cos \beta=-\frac{7}{25} \Rightarrow 2 \cos ^2 \frac{\beta}{2}-1=-\frac{7}{25} \\ & \Rightarrow \quad \cos ^2 \frac{\beta}{2}=\frac{9}{25} \Rightarrow \cos \frac{\beta}{2}=-\frac{3}{5} \\ & \Rightarrow \quad \sin \frac{\beta}{2}=\frac{4}{5} \Rightarrow \cot \frac{\beta}{2}=\frac{-3}{4} \\ & \therefore \sqrt{13} \sin \frac{\alpha}{2}+\cos \frac{\beta}{2}+\tan \frac{\alpha}{2} \cot \frac{\beta}{2} \\ & =\sqrt{13}\left(\frac{2}{\sqrt{13}}\right)+\left(-\frac{3}{5}\right)+\left(-\frac{2}{3}\right)\left(-\frac{3}{4}\right) \\ & =2-\frac{3}{5}+\frac{1}{2}=\frac{20-6+5}{10}=\frac{19}{10} \end{aligned} $

Given,

$ \begin{aligned} & \cos \frac{\pi}{7} \cos \frac{2 \pi}{7} \cos \frac{3 \pi}{7} \cos \frac{\pi}{14} \cos \frac{3 \pi}{14} \cos \frac{5 \pi}{14} \\ & =\frac{1}{8}\left[\begin{array}{l} \left(2 \cos \frac{5 \pi}{14} \cos \frac{\pi}{7}\right)\left(2 \cos \frac{3 \pi}{14} \cos \frac{2 \pi}{7}\right) \\ \left(2 \cos \frac{3 \pi}{7} \cos \frac{\pi}{14}\right) \end{array}\right] \end{aligned} $

$ =\frac{1}{8}\left[\begin{array}{l} \left.\cos \frac{7 \pi}{14}+\cos \frac{3 \pi}{14}\right)\left(\cos \frac{7 \pi}{14}+\cos \frac{\pi}{14}\right) \\ \left(\cos \frac{7 \pi}{14}+\cos \frac{5 \pi}{14}\right) \end{array}\right] $

$ \begin{aligned} & =\frac{1}{8}\left[\cos \frac{3 \pi}{14} \cos \frac{\pi}{14} \cos \frac{5 \pi}{14}\right] \\ & =\frac{1}{8}\left[\begin{array}{l} \cos \left(\frac{\pi}{2}-\frac{2 \pi}{7}\right) \cos \left(\frac{\pi}{2}-\frac{3 \pi}{7}\right) \\ \cos \left(\frac{\pi}{2}-\frac{\pi}{7}\right) \end{array}\right] \end{aligned} $

$ \begin{aligned} & =\frac{1}{8}\left[\sin \frac{2 \pi}{7} \sin \frac{3 \pi}{7} \sin \frac{\pi}{7}\right] \\ & =\frac{1}{16}\left[\left(2 \sin \frac{2 \pi}{7} \sin \frac{\pi}{7}\right) \sin \frac{3 \pi}{7}\right] \\ & =\frac{1}{16}\left[\left(-\cos \frac{3 \pi}{7}+\cos \frac{\pi}{7}\right) \sin \frac{3 \pi}{7}\right] \\ & =\frac{1}{16}\left[-\sin \frac{3 \pi}{7} \cos \frac{3 \pi}{7}+\sin \frac{3 \pi}{7} \cos \frac{\pi}{7}\right] \\ & =\frac{1}{32}\left[-2 \sin \frac{3 \pi}{7} \cos \frac{3 \pi}{7}+2 \sin \frac{3 \pi}{7} \cos \frac{\pi}{7}\right] \\ & =\frac{1}{32}\left[-\sin \frac{6 \pi}{7}+\sin \frac{4 \pi}{7}+\sin \frac{2 \pi}{7}\right] \\ & =\frac{1}{32}\left[-\sin \left(\pi-\frac{\pi}{7}\right)+\sin \left(\pi-\frac{3 \pi}{7}\right)+\sin \frac{2 \pi}{7}\right] \\ & =\frac{1}{32}\left[\sin \frac{2 \pi}{7}+\sin \frac{3 \pi}{7}-\sin \frac{\pi}{7}\right] \end{aligned} $

$\cot \theta=-\frac{2}{3}$

$\theta$ does not lie in 4th quadrant means $\theta$ lies in 2nd quadrant, then

$ \begin{aligned} & \sin \theta=\frac{3}{\sqrt{13}} \\ & \cos \theta=-\frac{2}{\sqrt{13}} \\ & \tan \theta=-\frac{3}{2} \end{aligned} $

$ \text { } \begin{aligned}Now, \frac{(5 \sin \theta+\cos \theta)^2}{\tan \theta+\cot \theta} & =\frac{\left(5 \cdot \frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}}\right)^2}{-\frac{3}{2}-\frac{2}{3}} \\ & =\frac{13}{-\frac{13}{6}}=-6 \end{aligned} $