Trigonometric Ratios & Identities

$ \begin{aligned} &\text { We have, }\\ &\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\tan ^3\left(\frac{\pi}{4}+\frac{\beta}{2}\right) \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \frac{1+\tan \frac{\alpha}{2}}{1-\tan \frac{\alpha}{2}}=\left(\frac{1+\tan \frac{\beta}{2}}{1-\tan \frac{\beta}{2}}\right)^3 \\ & \Rightarrow\left(\frac{\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}}\right)^2=\left[\left(\frac{\cos \frac{\beta}{2}+\sin \frac{\beta}{2}}{\cos \frac{\beta}{2}-\sin \frac{\beta}{2}}\right)^2\right]^3 \\ & \Rightarrow \frac{1+\sin \alpha}{1-\sin \alpha}=\left(\frac{1+\sin \beta}{1-\sin \beta}\right)^3 \\ & =\frac{1+\sin \alpha}{1-\sin \alpha}=\frac{1+3 \sin \beta+3 \sin \beta^2+\sin ^3 \beta}{1-3 \sin \beta+3 \sin ^2 \beta-\sin ^3 \beta} \end{aligned}\\ &\text { Apply componendo and dividendo }\\ &\begin{aligned} \frac{1}{\sin \alpha} & =\frac{1+3 \sin ^2 \beta}{3 \sin \beta+\sin ^3 \beta} \\ \Rightarrow \frac{3+\sin ^2 \beta}{1+3 \sin ^2 \beta} & =\frac{\sin \alpha}{\sin \beta} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } P=\sin \frac{2 \pi}{7}+\sin \frac{4 \pi}{7}+\sin \frac{8 \pi}{7} \\ & Q=\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{8 \pi}{7} \\ & P^2+Q^2=3+2\left(\cos \frac{2 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \frac{4 \pi}{7}\right) \\ & \text { Also, } \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=\frac{1}{2} \\ & \Rightarrow \quad P^2+Q^2=4 \Rightarrow \sqrt{P^2+Q^2}=2 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \tan ^2\left(\frac{\alpha}{2}\right) & =\frac{1-\cos \alpha}{1+\cos \alpha} \\ \tan ^2\left(\frac{\alpha}{2}\right) & =\frac{1-\left(\frac{l \cos \beta+m}{l+m \cos \beta}\right)}{1+\left(\frac{l \cos \beta+m}{l+m \cos \beta}\right)} \\ & =\frac{l(1-\cos \beta)-m(1-\cos \beta)}{l(1+\cos \beta)+m(1+\cos \beta)} \end{aligned} $

$ \begin{array}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\left(\frac{l-m}{l+m}\right) \tan ^2\left(\frac{\beta}{2}\right) \\ \Rightarrow \quad \frac{\tan ^2\left(\frac{\alpha}{2}\right)}{\tan ^2\left(\frac{\beta}{2}\right)}=\frac{l-m}{l+m} \end{array} $

$ \begin{aligned} & \text { }(\cos \theta+\sin \theta)^2=2 \cos ^2 \theta \\ & \qquad \begin{aligned} & \sin 2 \theta=\cos 2 \theta \\ & \text { So, sec }(2 \theta)+\tan (2 \theta) \\ &=\frac{1+\sin 2 \theta}{\cos 2 \theta}=\frac{1+\cos 2 \theta}{\sin 2 \theta} \\ &=\frac{2 \cos ^2 \theta}{2 \cos \theta \cdot \sin \theta}=\cot \theta \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We know that }\\ &\begin{aligned} \tanh (2 x) & =\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} \\ & =\frac{e^{4 x}-1}{e^{4 x}+1} \\ & =\frac{3^4-1}{3^4+1}=\frac{80}{244} \quad(x=\ln 3) \\ \operatorname{sech}(2 x) & =\frac{2}{e^{2 x}+e^{-2 x}}=\frac{2 e^{2 x}}{e^{4 x}+1} \\ & =\frac{2 \times 3^2}{3^4+1}=\frac{18}{244} \\ \tanh (2 x) & +\operatorname{sech}(2 x)=\frac{98}{82}=\frac{49}{41} \end{aligned} \end{aligned} $

$\sin A=\frac{-24}{25}, \cos B=\frac{15}{17}$

According to the question,

$A$ is 3rd quadrant, $B$ in 4th quadrant.

Now, $\sin (A+B)=\sin A \cos B+\cos A \sin B$

$ \begin{aligned} & \quad=\frac{-24}{25} \times \frac{15}{17}+\left(\frac{-7}{25}\right) \cdot\left(\frac{-8}{17}\right) \\ & \Rightarrow \quad \frac{8(-45+7)}{25(17)} \Rightarrow \frac{-37 \times 8}{25 \times 17} \\ & \therefore \sin (A+B) < 0 ; \pi < A < \frac{3 \pi}{2}, \frac{3 \pi}{2} < B < 2 \pi \\ & \frac{5 \pi}{2} < A+B < \frac{7 \pi}{2} \end{aligned} $

$\therefore A+B$ will be in 3rd quadrant.

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & 4 \cos \frac{7 \theta}{2} \cos \frac{3 \theta}{2} \sin 5 \theta \\ &= 2(\cos 5 \theta+\cos 2 \theta) \sin 5 \theta \\ & \quad(\because 2 \cos A \cos B=\cos (A+B) +\cos (A-B)) \\ &= 2 \sin 5 \theta \cos 5 \theta+2 \sin 5 \theta \cos 2 \theta \\ &= \sin 10 \theta+\sin 7 \theta+\sin 3 \theta \\ &(\because 2 \sin A \cos B=\sin (A+B)+\sin (A-B)) \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \operatorname{coth}^2 x-\tanh ^2 x \\ & =\frac{\cosh ^2 x}{\sinh ^2 x}-\frac{\sinh ^2 x}{\cosh ^2 x} \\ & =\frac{\left(\cosh ^2 x+\sinh ^2 x\right)\left(\cosh ^2 x-\sinh ^2 x\right)}{\sinh ^2 x \cdot \cosh ^2 x} \\ & =\frac{\left(\cosh ^2 x+\sinh ^2 x\right)}{\sinh ^2 x \cdot \cosh ^2 x}=\frac{4 \cosh 2 x}{(\sinh 2 x)^2} \\ & =4 \cosh 2 x(\operatorname{cosech} 2 x)^2 \end{aligned} $

$3 \sin \theta+4 \cos \theta=3$

On squaring both sides, we get $79 \sin ^2 \theta+16 \cos ^2 \theta+24 \sin \theta \cos \theta=9$

$ \begin{aligned} & \Rightarrow 9+7 \cos ^2 \theta+24 \sin \theta \cos \theta=9 \\ & \Rightarrow \cos \theta(7 \cos \theta+24 \sin \theta)=0 \\ & \Rightarrow \cos \theta \neq 0 \\ & \quad \theta \neq(2 n+1) \frac{\pi}{2} \\ & \text { and } 7 \cos \theta+24 \sin \theta=0 \\ & \begin{aligned} & \tan \theta=\frac{-7}{24} \\ & \therefore \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta} \\ & \quad= \frac{2 \times\left(\frac{-7}{24}\right)}{1+\frac{49}{576}}=\frac{-14 \times 24}{625} \\ & \quad=\frac{-336}{625} \end{aligned} \end{aligned} $

$ \text { } \frac{\cos 15^{\circ} \cos ^2 22 \frac{1}{2}^{\circ}-\sin 75^{\circ} \sin ^2 52 \frac{1}{2}^{\circ}}{\cos ^2 15^{\circ}-\cos ^2 75^{\circ}} $

$ =\frac{\cos 15^{\circ}\left(\cos ^2 22 \frac{1}{2}^{\circ}-\sin ^2 52 \frac{1}{2}^{\circ}\right)}{\cos ^2 15^{\circ}-\sin ^2 15^{\circ}}\binom{\because \sin 75^{\circ}=\sin \left(90^{\circ}-15^{\circ}\right)}{=\cos 15^{\circ}} $

$ \begin{aligned} & =\frac{\cos 15^{\circ}\left(\frac{1+\cos 45^{\circ}}{2}-\frac{1-\cos 105^{\circ}}{2}\right)}{\cos 30^{\circ}} \\ & =\frac{\cos 15^{\circ}}{2} \frac{\left[\cos 45^{\circ}+\cos \left(90+15^{\circ}\right)\right]}{\frac{\sqrt{3}}{2}} \\ & =\frac{1}{\sqrt{3}}\left(\cos 15^{\circ} \cos 45^{\circ}-\sin 15^{\circ} \cos 15^{\circ}\right) \\ & =\frac{1}{2 \sqrt{3}}\left(2 \cos 15^{\circ} \cos 45^{\circ}-2 \cos 15^{\circ} \sin 15^{\circ}\right) \\ & =\frac{1}{2 \sqrt{3}}\left[\cos \left(60^{\circ}\right)+\cos 30^{\circ}-\sin 30^{\circ}\right] \\ & =\frac{1}{2 \sqrt{3}}\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{4} \end{aligned} $

$ 16 \sin 12^{\circ} \cos 18^{\circ} \sin 48^{\circ} $

$ \begin{aligned} & =16 \sin 12^{\circ} \sin 72^{\circ} \sin 48^{\circ} \\ & =16 \sin 12^{\circ} \sin \left(60+12^{\circ}\right) \sin \left(60^{\circ}-12^{\circ}\right) \\ & =\frac{16}{4} \sin 3\left(12^{\circ}\right)=4 \sin 36^{\circ} \\ & =\frac{4 \sqrt{10-2 \sqrt{5}}}{4}=\sqrt{10-2 \sqrt{5}} \end{aligned} $

We have the expression: $5 \sin \theta + 3 \cos \left(\theta + \frac{\pi}{3}\right) + 3$

Step 1: Expand the cosine term

$3 \cos \left(\theta + \frac{\pi}{3}\right)$ can be written using the formula: $\cos(A+B) = \cos A \cos B - \sin A \sin B$.

$\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$.

So, $ 3 \cos \left(\theta + \frac{\pi}{3}\right) = 3 \left( \cos \theta \cdot \frac{1}{2} - \sin \theta \cdot \frac{\sqrt{3}}{2} \right) = \frac{3}{2} \cos \theta - \frac{3\sqrt{3}}{2} \sin \theta $

Step 2: Combine all terms

Now put this back into the main expression:

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

Combine the $\sin\theta$ terms:

$5 \sin \theta - \frac{3\sqrt{3}}{2} \sin \theta = \left(5 - \frac{3\sqrt{3}}{2}\right)\sin\theta$

So the expression becomes: $ \left(5 - \frac{3\sqrt{3}}{2}\right)\sin\theta + \frac{3}{2}\cos\theta + 3 $

Step 3: Find the maximum and minimum values

This is now in the form $a\sin\theta + b\cos\theta + c$, where $a = 5 - \frac{3\sqrt{3}}{2}$, $b = \frac{3}{2}$, $c = 3$.

The largest and smallest values of $a\sin\theta + b\cos\theta$ are $R$ and $-R$, where $R = \sqrt{a^2 + b^2}$.

Calculate $R$: $ R = \sqrt{\left(5 - \frac{3\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2} $

Expanding: $ = \sqrt{25 + \frac{27}{4} - 15\sqrt{3} + \frac{9}{4}} = \sqrt{25 - 15\sqrt{3} + 9} = \sqrt{34 - 15\sqrt{3}} $

Step 4: Substitute the extreme values

The minimum value is $-R + 3 = \alpha$. The maximum value is $R + 3 = \beta$.

So, $\alpha = -R + 3$ and $\beta = R + 3$.

Step 5: Calculate the final result

Find $\alpha + \beta$: $ \alpha + \beta = (-R + 3) + (R + 3) = 6 $

Find $\alpha - \beta$: $ \alpha - \beta = (-R + 3) - (R + 3) = -2R $

Now, $ (\alpha - \beta)(\alpha + \beta - 6) = (-2R) \times (6 - 6) = (-2R) \times 0 = 0 $

We have,

$ \begin{aligned} & \frac{\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 89^{\circ}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1} \\ & =\frac{\left(\sin 1^{\circ}+\sin 89^{\circ}\right)+\left(\sin 2^{\circ}+\sin 88^{\circ}\right)+\ldots+\left(\sin 44^{\circ}+\sin 46^{\circ}\right)+\sin 45^{\circ}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1} \\ & =\frac{2 \sin \frac{90^{\circ}}{2} \cos \frac{88^{\circ}}{2}+2 \sin \frac{90^{\circ}}{2} \cos \frac{86^{\circ}}{2}+\ldots+2 \sin \frac{90^{\circ}}{2} \cos \frac{2^{\circ}}{2}+\frac{1}{\sqrt{2}}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1} \\ & =\frac{\sqrt{2}\left(\cos 44^{\circ}+\cos 42^{\circ}+\ldots+\cos 2^{\circ}+\cos 1^{\circ}\right)+\frac{1}{\sqrt{2}}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1} \end{aligned} $

Let $x=\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}$

$ =\frac{\sqrt{2} x+\frac{1}{\sqrt{2}}}{2 x+1}=\frac{\frac{(2 x+1)}{\sqrt{2}}}{2 x+1}=\frac{1}{\sqrt{2}} $

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

$ \begin{aligned} \Rightarrow & 3[\sin \alpha \cos \beta-\cos \alpha \sin \beta] \\ & =5[\cos \alpha \cos \beta-\sin \alpha \sin \beta] \end{aligned} $

On dividing by $\cos \alpha \cos \beta$, we get

$ \begin{array}{ll} & 3[\tan \alpha-\tan \beta]=5(1-\tan \alpha \tan \beta) \\ \Rightarrow & 3 \tan \alpha-3 \tan \beta=5-5 \tan \alpha \tan \beta \\ \Rightarrow & 3 \tan \alpha+5 \tan \alpha \tan \beta=5+3 \tan \beta \\ \Rightarrow & \tan \alpha(3+5 \tan \beta)=5+3 \tan \beta \\ \Rightarrow & \tan \alpha=\frac{5+3 \tan \beta}{3+5 \tan \beta} \end{array} $

Now, $\tan \left(\frac{\pi}{4}-\alpha\right)=\frac{1-\tan \alpha}{1+\tan \alpha}$

And $\tan \left(\frac{\pi}{4}-\beta\right)=\frac{1-\tan \beta}{1+\tan \beta}$

$ \begin{aligned} & \therefore \quad \frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)}=\frac{(1-\tan \alpha)(1+\tan \beta)}{(1+\tan \alpha)(1-\tan \beta)} \\ & \therefore \quad 1-\tan \alpha=1-\frac{5+3 \tan \beta}{3+5 \tan \beta}=\frac{2(\tan \beta-1)}{3+5 \tan \beta} \\ & \text { And } \quad 1+\tan \beta=1+\frac{5+3 \tan \beta}{3+5 \tan \beta}=\frac{8(1+\tan \beta)}{3+5 \tan \beta} \\ & \therefore \quad \frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)}=\frac{\frac{2(\tan \beta-1)}{3+5 \tan \beta} \times(1+\tan \beta)}{8(1+\tan \beta)} \times(1-\tan \beta) \\ & \quad=\frac{-2}{8}=\frac{-1}{4} \end{aligned} $

Given, $\sin A=-\frac{60}{61}, \cot B=-\frac{40}{9}$

Now, $\cos ^2 A=1-\sin ^2 A$

$ \begin{aligned} & \Rightarrow \quad 1-\left(-\frac{60}{61}\right)^2 \Rightarrow 1-\frac{3600}{3721}=\frac{121}{3721} \\ & \therefore \quad \cos A= \pm \frac{11}{61} \end{aligned} $

But $\sin A<0$ and $A$ and $B$ are not in 4th quadrant.

So, $A$ in third quadrant and $B$ in 2nd quadrant

$ \therefore \quad \cos A=-\frac{11}{61} $

And $\cot A=\frac{\cos A}{\sin A}=\frac{-11}{-60}=\frac{11}{60}$

$ 6 \cot A=\frac{66}{60}=\frac{11}{10} $

Now, $\tan B=\frac{1}{\cot B}=\frac{-9}{40}$

And, $\sec ^2 B=1+\tan ^2 B$

$ \begin{aligned} & \Rightarrow \quad 1+\left(\frac{-9}{40}\right)^2=1+\frac{81}{1600}=\frac{41}{40} \\ & \Rightarrow \quad \sec B=-\frac{41}{40}(\because B \text { in 2nd quadrant }) \\ & \Rightarrow \quad 4 \sec B=\frac{-164}{40}=-\frac{41}{40} \\ & \therefore \quad 6 \cot A+4 \sec B=\frac{11}{10}-\frac{41}{10} \\ & \Rightarrow \quad \frac{-30}{10}=-3 \end{aligned} $

Given,

$ f(x)=\frac{2 \sin \left(\frac{\pi x}{3}\right) \cos \left(\frac{2 \pi x}{5}\right)}{3 \tan \left(\frac{7 \pi x}{2}\right)-5 \sec \left(\frac{5 \pi x}{3}\right)} $

We know that

$ \begin{aligned} & \sin (k x) \rightarrow \text { period }=\frac{2 \pi}{k} \\ & \cos (k x) \rightarrow \text { period }=\frac{2 \pi}{k} \\ & \tan (k x) \rightarrow \text { period }=\frac{\pi}{k} \\ & \sec (k x) \rightarrow \text { period }=\frac{2 \pi}{k} \end{aligned} $

So, for $\sin \left(\frac{\pi x}{3}\right)$, period $=\frac{2 \pi}{\frac{\pi}{3}}=6$

$ \begin{aligned} & \cos \left(\frac{2 \pi x}{5}\right) \text { period }=\frac{2 \pi}{\frac{2 \pi}{5}}=5 \\ & \tan \left(\frac{7 \pi x}{2}\right) \text { period }=\frac{\pi}{\frac{7 \pi}{2}}=\frac{2}{7} \\ & \sec \left(\frac{5 \pi x}{3}\right) \text { period }=\frac{2 \pi}{\frac{5 \pi}{3}}=\frac{6}{5} \end{aligned} $

Now, the LCM of $6,5, \frac{2}{7}$ and $\frac{6}{5}$ is as follows

$\operatorname{LCM}$ of $\left\{\frac{6}{1}, \frac{5}{1}, \frac{2}{7}, \frac{6}{5}\right\}=\frac{\operatorname{LCM} \text { of }(6,5,2,6)}{\text { HCF of }(1,1,7,5)}$

$ \Rightarrow \frac{30}{1}=30 $

So, the period of $f(x)$ is 30 .

$ \begin{aligned} &\text { Given, } A+B+C=4 S\\ &\begin{aligned} & \therefore A+B+C=4 S=\pi \Rightarrow S=\frac{\pi}{4} \\ & \therefore \sin (2 S-A)+\sin (2 S-B)+\sin (2 S-C)-\sin 2 S \\ & =\sin \left(2 \times \frac{\pi}{4}-A\right)+\sin \left(2 \times \frac{\pi}{4}-B\right) +\sin \left(2 \times \frac{\pi}{4}-C\right)-\sin \left(2 \times \frac{\pi}{4}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \begin{aligned} = & \sin \left(\frac{\pi}{2}-A\right)+\sin \left(\frac{\pi}{2}-B\right)+\sin \left(\frac{\pi}{2}-C\right)-\sin \left(\frac{\pi}{2}\right) \end{aligned} \\ & =\cos A+\cos B+\cos C-1 \\ & =4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \end{aligned} $

Let $f(x)=\sec (x)$

Since $58^{\circ}$ is close to $60^{\circ}$ and $\sec \left(60^{\circ}\right)=2$

Also, $60^{\circ}=60 \times 0.0175$ radians $=1.05$

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

$ \begin{aligned} \Rightarrow \quad f^{\prime}(x) & =\sec (x) \tan (x) \\ f^{\prime}\left(60^{\circ}\right) & =\sec \left(60^{\circ}\right) \tan \left(60^{\circ}\right) \\ & =2 \times \sqrt{3}=2 \sqrt{3} \end{aligned} $

And, change in angle,

$ \begin{aligned} \Delta x & =58^{\circ}-68^{\circ}=-2^{\circ} \\ & =-2 \times 0.0175 \text { radians } \\ & =-0.035 \text { radians } \end{aligned} $

Now, $\Delta y \approx f^{\prime}(x) \cdot \Delta x$

$ \begin{aligned} & \approx 2 \sqrt{3} \times(-0.035) \\ & \approx 2 \times 1.732 \times(-0.035) \\ & \approx-0.12124 \end{aligned} $

So, $\sec \left(58^{\circ}\right) \approx f\left(60^{\circ}\right)+\Delta y$

$ \Rightarrow \quad 2-0.12124 \approx 1.87876 $

So, the approximate value of $\sec \left(58^{\circ}\right)$ is 1.8788 .

$ \begin{array}{r} \text { }\sin \frac{\pi}{12} \cdot \sin \frac{2 \pi}{12} \cdot \sin \frac{3 \pi}{12} \cdot \sin \frac{4 \pi}{12} \cdot \sin \frac{5 \pi}{12} \cdot \sin \frac{6 \pi}{12} \\ =\sin 15^{\circ} \cdot \sin 30^{\circ} \cdot \sin 45^{\circ} \cdot \sin 60^{\circ} \sin 75^{\circ} \cdot \sin 90^{\circ} \end{array} $

$ \begin{aligned} & =\frac{\sqrt{6}-\sqrt{2}}{4} \times \frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} \times \frac{\sqrt{6}+\sqrt{2}}{4} \times 1 \\ & =\frac{(\sqrt{6})^2-(\sqrt{2})^2}{4 \times 4} \times \frac{\sqrt{3}}{4 \sqrt{2}} \\ & =\frac{1}{4} \times \frac{\sqrt{3}}{4 \sqrt{2}}=\frac{\sqrt{3}}{16 \sqrt{2}} \end{aligned} $

$ \begin{aligned} & \text { }(\sin A+\sin D)+(\sin B+\sin C) \\ & \qquad \begin{aligned} =2 \sin \left(\frac{A+D}{2}\right) & \cos \left(\frac{A-D}{2}\right) + 2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right) \end{aligned} \end{aligned} $

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

$ \begin{array}{r} =2 \sin \left(\frac{A+D}{2}\right)\left[\cos \left(\frac{A-D}{2}\right)+\cos \left(\frac{B-C}{2}\right)\right.] \\ =2 \sin \left(\frac{A+D}{2}\right)\left[2 \cos \left(\frac{\left(\frac{A-D+B-C}{2}\right)}{2}\right)\right. \left.\cos \left(\frac{\left(\frac{A-D-B+C}{2}\right)}{2}\right)\right] \end{array} $

$ \begin{array}{r} =4 \sin \left(\frac{A+D}{2}\right)\left[\cos \left(\frac{A+B-(C+D)}{4}\right)\right. \left.\cos \left(\frac{A+C-(B+D)}{4}\right)\right] \end{array} $

$ \begin{aligned} & \because \quad C+D=2 \pi-(A+B) \\ & \text { and } B+D=2 \pi-(A+C) \\ & =4 \sin \left(\frac{A+D}{2}\right)\left[\cos \left(\frac{\pi}{2}-\left(\frac{A+B}{2}\right)\right)\right. \left. \cos \left(\frac{\pi}{2}-\left(\frac{A+C}{2}\right)\right)\right] \\ & =4 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A+C}{2}\right) \sin \left(\frac{A+D}{2}\right) \end{aligned} $

$ \begin{aligned} &\text { } \cos x+\sin x=\frac{1}{2}, 0 < x < \pi\\ &\begin{aligned} \Rightarrow & \quad(\cos x+\sin x)^2=\frac{1}{4} \\ \Rightarrow & \quad 1+\sin 2 x=\frac{1}{4} \\ \Rightarrow & \quad \sin 2 x=\frac{-3}{4} \\ \Rightarrow & \quad \frac{2 \tan x}{1+\tan ^2 x}=\frac{-3}{4} \\ \Rightarrow & \quad 8 \tan x=-3-3 \tan ^2 x \\ \Rightarrow \quad & \quad 3 \tan ^2 x+8 \tan x+3=0 \\ \Rightarrow \quad & \tan x=\frac{-8 \pm \sqrt{64-36}}{6} \\ & =\frac{-8 \pm \sqrt{28}}{6} \\ & =\frac{-4 \pm \sqrt{7}}{3} \\ & =\frac{-4-\sqrt{7}}{3}, \frac{-4+\sqrt{7}}{3} \end{aligned} \end{aligned} $

Given, $\sin \theta+2 \cos \theta=1$

$ \begin{aligned} \Rightarrow & 2 \cos \theta=1-\sin \theta \\ \Rightarrow & 4 \cos ^2 \theta=1+\sin ^2 \theta-2 \sin \theta \\ \Rightarrow & \sin ^2 \theta+1-4\left(1-\sin ^2 \theta\right)-2 \sin \theta=0 \\ \Rightarrow & 5 \sin ^2 \theta-2 \sin \theta-3=0 \\ \Rightarrow & 5 \sin ^2 \theta-5 \sin \theta+3 \sin \theta-3=0 \\ \Rightarrow & 5 \sin \theta(\sin \theta-1)+3(\sin \theta-1)=0 \\ & (\sin \theta-1)(5 \sin \theta+3)=0 \\ & \sin \theta=1, \quad \sin \theta=\frac{-3}{5} \end{aligned} $

$\because \theta$ lies in 4th quadrant

$ \begin{aligned} & \therefore \sin \theta=\frac{-3}{5} \text { and } \cos \theta=\frac{4}{5} \\ & \therefore 7 \cos \theta+6 \sin \theta=\frac{28}{5}-\frac{18}{5}=\frac{10}{5}=2 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } 3 \cos ^2 A+2 \cos ^2 B=4 \\ & \Rightarrow \quad \frac{3(1+\cos 2 A)}{2}+\frac{2(1+\cos 2 \beta)}{2}=4 \\ & \Rightarrow \quad 3 \cos 2 A+2 \cos 2 \beta=3 \\ & \Rightarrow \quad 2 \cos 2 \beta=3(1-\cos 2 A)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { and } \frac{3 \sin A}{\sin B}=\frac{2 \cos B}{\cos A} \\ & \Rightarrow \quad 3 \sin A \cos A=2 \sin B \cos B \\ & \Rightarrow \quad \frac{3}{2} \sin 2 A=\sin 2 B \\ & \Rightarrow \quad 3 \sin 2 A=2 \sin 2 B \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 9 \sin ^2 2 A=4 \sin ^2 2 B \\ & \Rightarrow \quad 9-9 \cos ^2 2 A=4-4 \cos ^2 2 B \\ & 9 \cos ^2 2 A-4 \cos ^2 2 B=5 \end{aligned} $

From Eq. (i)

$ \begin{aligned} & \Rightarrow 9 \cos ^2 2 A-9(1-\cos 2 A)^2=5 \\ & \Rightarrow 9 \cos ^2 2 A-9 \cos ^2 2 A-9+18 \cos 2 A=5 \\ & \Rightarrow 18 \cos 2 A=14 \\ & \Rightarrow \cos 2 A=\frac{14}{18}=\frac{7}{9} \end{aligned} $

From Eq. (i), we get

$ \begin{aligned} \therefore \cos 2 B & =\frac{3}{2}\left(1-\frac{14}{18}\right), \cos A=\sqrt{\frac{1+\cos 2 A}{2}} \\ & =\frac{3}{2} \times \frac{4}{18}=\frac{1}{3} \quad=\sqrt{\frac{16}{18}}=\frac{2 \sqrt{2}}{3} \end{aligned} $

Now,

$ \begin{aligned} \sin (A+2 B) & =\sin A \cos 2 B+\cos A \sin 2 B \\ & =\frac{1}{3} \times \frac{1}{3}+\frac{2 \sqrt{2}}{3} \times \frac{2 \sqrt{2}}{3} \\ & =\frac{1}{9}+\frac{8}{9}=1 \\ \therefore A+2 B & =\frac{\pi}{2} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \cos ^2 \frac{3 \pi}{20}-1\right) \\ & \left(4 \cos ^2 \frac{5 \pi}{20}+1\right)\left(4 \cos ^2 \frac{7 \pi}{20}-1\right) \left(4 \cos ^2 \frac{9 \pi}{20}-1\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & =\left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \cos ^2 \frac{3 \pi}{20}-1\right)\left(4 \times \frac{1}{2}+1\right) \\ & \left(4 \cos ^2 \frac{7 \pi}{20}-1\right)\left(4 \cos ^2 \frac{9 \pi}{20}-1\right) \\ & =3\left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \cos ^2 \frac{9 \pi}{20}-1\right) \\ & \left(4 \cos ^2 \frac{3 \pi}{20}-1\right)\left(4 \cos ^2 \frac{7 \pi}{20}-1\right) \\ & =3\left(4 \cos ^2 \frac{\pi}{20}-1\right)\left(4 \sin ^2 \frac{\pi}{20}-1\right) \\ & \left(4 \cos ^2 \frac{3 \pi}{20}-1\right)\left(4 \sin ^2 \frac{3 \pi}{20}-1\right) \\ & =3\left(16 \cos ^2 \frac{\pi}{20} \cdot \sin ^2 \frac{\pi}{20}-4+1\right) \left(16 \sin ^2 \frac{3 \pi}{20} \cos ^2 \frac{3 \pi}{20}-4-1\right) \end{aligned} $

$ \begin{aligned} & =3\left(4 \sin ^2 \frac{\pi}{10}-3\right)\left(4 \sin ^2 \frac{3 \pi}{10}-3\right) \\ & =\frac{3}{\sin \frac{\pi}{10} \cdot \sin \frac{3 \pi}{10}}\left(4 \sin ^3 \frac{\pi}{10}-3 \sin \frac{\pi}{10}\right) \left(4 \sin ^3 \frac{3 \pi}{10}-3 \sin \frac{3 \pi}{10}\right) \end{aligned} $

$ \begin{aligned} & =\frac{3}{\sin \frac{\pi}{10} \sin \frac{3 \pi}{10}} \cdot \sin \frac{3 \pi}{10} \cdot \sin \left(\frac{9 \pi}{10}\right) \\ & =3 \quad\left(\because \sin \left(\frac{9 \pi}{10}\right)=\sin \left(\pi-\frac{\pi}{10}\right)=\sin \frac{\pi}{10}\right) \end{aligned} $

$ \begin{aligned} &\text {Given, }\\ &\begin{aligned} & 2 \tan (A+B)=3 \tan (A-B) \\ & \Rightarrow \frac{\tan (A+B)}{\tan (A-B)}=\frac{3}{2} \\ & \Rightarrow \frac{\tan (A+B)+\tan (A-B)}{\tan (A+B)-\tan (A-B)}=\frac{3+2}{3-2} \\ & \Rightarrow \frac{[\sin (A+B) \cos (A-B)}{[\sin (A+B) \cos (A-B)} -\sin (A-B) \cos (A+B)] \\ & \Rightarrow \frac{\sin (A+B+A-B)}{\sin (A+B-A+B)}=5 \\ & \Rightarrow \sin 2 A=5 \sin 2 B \\ & \Rightarrow \sin A \cos A=5 \sin B \cos B \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \cos ^3 80^{\circ}+\cos ^3 40^{\circ}-\cos ^3 20^{\circ}=k \\ & \left(\cos 80^{\circ}+\cos 40^{\circ}\right)\left(\left(\cos 80^{\circ}+\cos 40^{\circ}\right)^2\right. \left.-3 \cos 80^{\circ} \cos 40^{\circ}\right)-\cos ^3 20^{\circ}=k \\ & \Rightarrow \quad 2 \cos 60^{\circ} \cos 20^{\circ}\left(\left(2 \cos 60^{\circ} \cos 20^{\circ}\right)^2\right. \left.-\frac{3}{2}\left(\cos 120^{\circ}+\cos 40^{\circ}\right)\right]-\cos ^3 20^{\circ}=k \\ & \Rightarrow \quad \cos 20^{\circ}\left(\cos ^2 20^{\circ}-\frac{3}{2}\left(-\frac{1}{2}+\cos 40^{\circ}\right)\right. \left.-\cos ^3 20^{\circ}=k\right) \\ & \Rightarrow \quad \cos ^3 20^{\circ}-\frac{3}{2} \cos 20^{\circ}\left(-\frac{1}{2}+\cos 40^{\circ}\right) -\cos ^3 20^{\circ}=k \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{3}{4} \cos 20^{\circ}-\frac{3}{2} \cos 20^{\circ} \cos 40^{\circ}=k \\ & \Rightarrow \quad \frac{3}{4} \cos 20^{\circ}-\frac{3}{4}\left(\cos 60^{\circ}+\cos 20^{\circ}\right)=k \\ & \Rightarrow \quad \frac{3}{4} \cos 20-\frac{3}{4} \times \frac{1}{2}-\frac{3}{4} \cos 20^{\circ}=k \\ & \Rightarrow \quad k=-\frac{3}{8} \Rightarrow \frac{4 k}{3}=\frac{-1}{2}=\cos \frac{2 \pi}{3} \end{aligned} $

$ \begin{aligned} & \text { } \begin{array}{l} \cos 13^{\circ} \cdot \sin 17^{\circ} \cdot \sin 21^{\circ} \cdot \cos 47^{\circ} \\ =\left(\cos 13^{\circ} \cos 47^{\circ}\right) \cdot\left(\sin 17^{\circ} \sin 21^{\circ}\right) \\ =\frac{1}{2}\left[\cos \left(34^{\circ}\right)+\cos \left(60^{\circ}\right)\right] \cdot \frac{1}{2}\left[\cos \left(4^{\circ}\right)-\cos \left(38^{\circ}\right)\right] \\ \quad\left[\because \sin A \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)\right. \text { and } \cos A \cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)] \\ =\frac{1}{4}\left[\cos 34^{\circ}+\frac{1}{2}\right]\left[\cos 4^{\circ}-\cos 38^{\circ}\right] \\ =\frac{1}{4}\left[\cos 34^{\circ} \cos 4^{\circ}-\cos 34^{\circ} \cos 38^{\circ}+\frac{1}{2} \cos 4^{\circ}-\frac{1}{2} \cos 38^{\circ}\right] \end{array} \end{aligned} $

$ \begin{aligned} & =\frac{1}{4}\left[\frac{1}{2}\left\{\cos \left(30^{\circ}\right)+\cos \left(38^{\circ}\right)\right\}-\frac{1}{2}\left\{\cos \left(4^{\circ}\right)+\cos \left(72^{\circ}\right)\right\}\right. \left.+\frac{1}{2} \cos 4^{\circ}-\frac{1}{2} \cos 38^{\circ}\right] \\ & =\frac{1}{4}\left[\frac{1}{2}\left(\frac{\sqrt{3}}{2}+\cos 38^{\circ}\right)-\frac{1}{2}\left(\cos 4^{\circ}+\cos 72^{\circ}\right)+\frac{1}{2} \cos 4^{\circ}-\frac{1}{2} \cos 38^{\circ}\right] \\ & =\frac{1}{8}\left[\frac{\sqrt{3}}{2}-\cos 72^{\circ}\right]=\frac{\sqrt{3}}{16}-\frac{\cos 72^{\circ}}{8} \\ & =\frac{\sqrt{3}}{16}-\frac{\sqrt{5}-1}{4} \times \frac{1}{8} \left(\because \cos 72^{\circ}=\frac{\sqrt{5}-1}{4}\right)\\ & =\frac{\sqrt{3}}{16}-\frac{\sqrt{5}-1}{32}=\frac{2 \sqrt{3}-\sqrt{5}+1}{32}=\frac{1}{32}(2 \sqrt{3}-\sqrt{5}+1) \end{aligned} $

$ \begin{aligned} & \sin \frac{\pi}{5}+\sin \frac{2 \pi}{5}+\sin \frac{3 \pi}{5}+\sin \frac{4 \pi}{5} \\ & =\sin \frac{\pi}{5}+\sin \frac{2 \pi}{5}+\sin \left(\pi-\frac{2 \pi}{5}\right)+\sin \left(\pi-\frac{\pi}{5}\right) \\ & =\sin \frac{\pi}{5}+\sin \frac{2 \pi}{5}+\sin \frac{2 \pi}{5}+\sin \frac{\pi}{5}=2 \sin \frac{\pi}{5}+2 \sin \frac{2 \pi}{5} \\ & =2\left[\sin \frac{\pi}{5}+\sin \frac{2 \pi}{5}\right]=4 \sin \frac{3 \pi}{10} \cos \frac{\pi}{10} \\ & =4 \frac{(\sqrt{5}+1)}{4} \frac{(\sqrt{10+2 \sqrt{5}})}{4}=\frac{1}{4}(\sqrt{5}+1)(\sqrt{10+2 \sqrt{5}}) \end{aligned} $

$ \begin{aligned} \operatorname{cosec} 48^{\circ}+\operatorname{cosec} 96^{\circ} & +\operatorname{cosec} 192^{\circ} +\operatorname{cosec} 384^{\circ} \end{aligned} $

$ \begin{aligned} \because \quad \operatorname{cosec} \theta & =\cot \theta / 2-\cot \theta \\ \therefore \operatorname{cosec} 48^{\circ} & =\cot 24^{\circ}-\cot 48^{\circ} \\ \operatorname{cosec} 96^{\circ} & =\cot 48^{\circ}-\cot 96^{\circ} \\ \operatorname{cosec} 192^{\circ} & =\cot 96^{\circ}-\cot 192^{\circ} \\ \operatorname{cosec} 384^{\circ} & =\operatorname{cosec} 24^{\circ} \\ =\cot 24^{\circ} & -\cot 48^{\circ}+\cot 48^{\circ}-\cot 96^{\circ}+\cot 96^{\circ}-\cot 192^{\circ}+\cos 24^{\circ} \end{aligned} $

$ \begin{aligned} &=\cot 24^{\circ}-\cot 192^{\circ}+\cos 24^{\circ}\\ &\text { and } \operatorname{cosec} 24^{\circ}=\cot 12^{\circ}-\cot 24^{\circ}\\ &\begin{aligned} & =\cot 24^{\circ}-\cot 192^{\circ}+\cot 12^{\circ}-\cot 24^{\circ} \\ & =\cot 12^{\circ}-\cot 192^{\circ} \\ & =\cot 12^{\circ}-\cot \left(180+12^{\circ}\right) \\ & =\cot 12^{\circ}-\cot 12^{\circ}=0 \end{aligned} \end{aligned} $

$\because \cos \theta=-\frac{3}{5}<0$ and $\theta$ does not lie in second quadrant.

Thus, $\theta$ lies is third quadrant.

So, $\theta / 2$ must be lie in second quadrant.

There $\tan \theta / 2$ must be negative.

then, $\cos \theta=\frac{3}{5}=\frac{B}{H}$

$\therefore \quad \tan \theta=\frac{4}{3}$

So, $\tan \theta=\frac{2 \tan \theta / 2}{1-\tan ^2 \theta / 2}$

let $\tan \theta / 2=x$

$ \begin{aligned} & \Rightarrow \quad \frac{4}{3}=\frac{2 x}{1-x^2} \Rightarrow 4-4 x^2=6 x \\ & \Rightarrow \quad 2 x^2+3 x-2=0 \\ & \Rightarrow \quad 2 x^2+4 x-x-2=0 \\ & \Rightarrow \quad(2 x-1)(x+2)=0 \end{aligned} $

Therefore,

$ \begin{aligned} & x=1 / 2 \text { or } x=-2 \\ \Rightarrow \quad & \tan \theta / 2=1 / 2 \text { or } \tan \theta / 2=-2 \end{aligned} $

$\because \tan \theta / 2$ is negative

Hence, $\tan \theta / 2=-2$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \begin{aligned} Let\,\,f(x) & =\cos ^2\left(\frac{x}{4}\right)+\sin \left(\frac{x}{4}\right) \\ & =1-\sin ^2\left(\frac{x}{4}\right)+\sin \left(\frac{x}{4}\right) \\ & =-\sin ^2\left(\frac{x}{4}\right)+\sin \left(\frac{x}{4}\right)+1 \\ & =\frac{5}{4}-\left(\sin \left(\frac{x}{4}\right)-\frac{1}{2}\right)^2 \end{aligned} \\ & \left.\therefore f(x)\right|_{\max }=\alpha=\frac{5}{4} \text { when } \sin \frac{x}{4}=\frac{1}{2} \\ & \text { and }\left.f(x)\right|_{\min }=\beta=-1 \text { when } \sin \frac{x}{4}=-1 \\ & \therefore \alpha-\beta=\frac{5}{4}-(-1)=\frac{9}{4} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } 3 \cos ^2 A+2 \cos ^2 B=4 \\ & \Rightarrow 3\left(1-\sin ^2 A\right)+2\left(1-\sin ^2 B\right)=4 \end{aligned} $

Now, $3 \sin A \cdot \cos A=2 \sin B \cdot \cos B$ (given)

$ \Rightarrow \quad 3 \sin 2 A=2 \sin 2 B \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & \cos (A+2 B)=\cos A \cos 2 B-\sin A \sin 2 B \\ & =\cos A \cdot 3 \sin ^2 A-\sin A \cdot \frac{3}{2} \cdot 2 \sin A \cos A \,\,\,\,(from Eq. (i))\text { } \end{aligned} $

$ =3 \cos A \sin ^2 A-3 \cos A \sin ^2 A=0 $

$ \therefore A+2 B=90^{\circ} $

$ \begin{aligned} &\text { Given, } \sin x-\sin y=27 / 65 \text { and }\\ &\begin{aligned} & \cos x-\cos y=\frac{-21}{65} \\ & 2 \sin \left(\frac{x-y}{2}\right) \cos \left(\frac{x+y}{2}\right)=\frac{27}{65} \text { and } \\ & 2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)=\frac{21}{65} \\ & \therefore \tan \left(\frac{x+y}{2}\right)=\frac{21}{27}=\frac{7}{9} \\ & \sin (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^2 \frac{x+y}{2}}=\frac{2 \times \frac{7}{9}}{1+\frac{49}{81}} \\ & \sin (x+y)=\frac{63}{65} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \sin \beta=\frac{5}{6} \sin \alpha\\ &\begin{aligned} & \Rightarrow \cos \beta=\frac{5 \sqrt{5}}{9} \cos \alpha \\ & \Rightarrow \sin ^2 \beta+\cos ^2 \beta=1 \\ & \Rightarrow\left(\frac{5}{6} \sin \alpha\right)^2+\left(\frac{5 \sqrt{5}}{9} \cos \alpha\right)^2=1 \\ & \Rightarrow \frac{25}{36} \sin ^2 \alpha+\frac{125}{81} \cos ^2 \alpha=1 \\ & \Rightarrow \frac{25}{36} \sin ^2 \alpha+\frac{125}{81}-\frac{125}{81} \sin ^2 \alpha=1 \\ & \Rightarrow\left(\frac{225-500}{324}\right) \sin ^2 \alpha=\frac{81-125}{81} \\ & \Rightarrow \sin ^2 \alpha=\frac{-44}{81} \times \frac{324}{-275} \\ & \Rightarrow \sin ^2 \alpha=\frac{4}{9} \times \frac{11}{9} \times \frac{4}{25} \times \frac{81}{11} \\ & \Rightarrow \sin ^2 \alpha=\frac{16}{25} \Rightarrow \sin \alpha=\frac{4}{5} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { }\left(\frac{\sin 3 \theta}{\sin \theta}\right)^2-\left(\frac{\cos 3 \theta}{\cos \theta}\right)^2=a \cos b \theta \\ & \Rightarrow\left(\frac{3 \sin \theta-4 \sin ^3 \theta}{\sin \theta}\right)-\left(\frac{4 \cos ^3 \theta-3 \cos \theta}{\cos \theta}\right) =a \cos b \theta \\ & \Rightarrow\left(3-4 \sin ^2 \theta\right)^2-\left(4 \cos ^2 \theta-3\right)^2 =a \cos b \theta \\ & \Rightarrow 9+16 \sin ^4 \theta-24 \sin ^2 \theta-16 \cos ^4 \theta -9+24 \cos ^2 \theta=a \cos b \theta \\ & \Rightarrow 24\left(\cos ^2 \theta-\sin ^2 \theta\right)-16\left(\cos ^4 \theta-\sin ^4 \theta\right) =a \cos ^b \theta \\ & \Rightarrow 8\left(\cos ^2 \theta-\sin ^2 \theta\right)\left(3-2\left(\cos ^2 \theta+\sin ^2 \theta\right)\right) =a \cos b \theta \quad, \\ & \Rightarrow 8 \cos 2 \theta(3-2)=a \cos b \theta \\ & \Rightarrow 8 \cos 2 \theta=a \cos b \theta \Rightarrow a=8, b=2 \\ & \therefore a: b=8: 2=4: 1 \end{aligned} $

Let $x_1$ be the initial horizontal distance

$ \tan 15^{\circ}=\frac{5}{x_1} \Rightarrow x_1=\frac{5}{\tan 15^{\circ}} $

Let $x_2$ be the final horizontal distance

$ \begin{array}{rlrl} \tan 30^{\circ} & =\frac{5}{x_2} \\ \Rightarrow & \quad x_2 & =\frac{5}{\tan 30^{\circ}} \end{array} $

The distance travelled is $d=x_1-x_2$

$ \begin{aligned} & d=\frac{5}{\tan 15^{\circ}}-\frac{5}{\tan 30^{\circ}} \\ & d \approx 5(3.732-1.732) \\ & d \approx 5 \times 2 \\ & d=10 \mathrm{~km} \end{aligned} $

$ \begin{aligned} &\begin{aligned} \text { speed } & =\frac{d}{t}=\frac{10}{50}=0.2 \mathrm{~km} / \mathrm{s} \\ & =3600 \times 0.2 \mathrm{~km} / \mathrm{h}=720 \mathrm{~km} / \mathrm{h} \end{aligned}\\ &\Rightarrow \text { The speed of the plane is } 720 \mathrm{~km} / \mathrm{h} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} A+B= & \frac{\pi}{4} \\ \Rightarrow \frac{\cos B-\sin B}{\cos B+\sin B} & =\frac{\sin \left(\frac{\pi}{4}+A\right)-\sin \left(\frac{\pi}{4}-A\right)}{\sin \left(\frac{\pi}{4}+A\right)+\sin \left(\frac{\pi}{4}-A\right)} \\ & =\frac{2 \cos \frac{\pi}{4} \sin A}{2 \sin \frac{\pi}{4} \cos A}=\tan A \end{aligned} \end{aligned} $

$7 \cos \theta-\sin \theta=5, \tan \theta>0$

$ 7-\tan \theta=5 \sec \theta $

Squaring both sides, we get

$ \begin{aligned} & \Rightarrow 49+\tan ^2 \theta-14 \tan \theta=25\left(1+\tan ^2 \theta\right) \\ & \Rightarrow 24 \tan ^2 \theta+14 \tan \theta-24=0 \\ & \Rightarrow 12 \tan ^2 \theta+7 \tan \theta-12=0 \\ & \Rightarrow 12 \tan ^2 \theta+16 \tan \theta-9 \cdot \tan \theta-12=0 \\ & \Rightarrow(3 \tan \theta+4)(4 \tan \theta-3)=0 \\ & \quad \tan \theta=\frac{3}{4} \end{aligned} $

$ \begin{aligned} &\text { } \sin ^3 10^{\circ}+\sin ^3 50^{\circ}-\sin ^3 70^{\circ}\\ &\begin{aligned} & \Rightarrow \quad \sin 3 x=3 \sin x-4 \sin ^3 x \\ & \Rightarrow \quad \sin ^3 x=\frac{1}{4}(3 \sin x-\sin 3 x) \\ & \Rightarrow \quad \frac{1}{4}\left[3 \sin 10^{\circ}-\sin 30^{\circ}+3 \sin 50^{\circ}\right. \left.-\sin 150^{\circ}-3 \sin 70^{\circ}+\sin 210^{\circ}\right] \\ & \Rightarrow \quad \frac{1}{4}\left[3 \sin 10^{\circ}+3 \sin 50^{\circ}-3 \sin 70^{\circ}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right] \\ & \Rightarrow \frac{3}{4}\left[\sin 10^{\circ}+\sin 50^{\circ}-\sin 70^{\circ}-\frac{1}{2}\right] \\ & \Rightarrow \frac{3}{4}\left[2 \sin 30^{\circ} \cos 20^{\circ}-\sin 70^{\circ}-\frac{1}{2}\right] \\ & \Rightarrow \frac{3}{4}\left[\cos 20^{\circ}-\sin 70^{\circ}-\frac{1}{2}\right] \\ & \Rightarrow \frac{3}{4}\left[\sin 70^{\circ}-\sin 70^{\circ}-\frac{1}{2}\right]=-\frac{3}{8} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, } \sum_{r=1}^{89} \frac{1}{\sin r^{\circ} \cdot \sin (r+1)^{\circ}}\\ &\begin{aligned} & \Rightarrow \frac{1}{\sin 1^{\circ}} \sum_{r=1}^{89} \frac{\sin \left(\left(r^{\circ}+1^{\circ}\right)-r^{\circ}\right)}{\sin r^{\circ} \cdot \sin \left(r^{\circ}+1^{\circ}\right)} \\ & \Rightarrow \frac{1}{\sin 1^{\circ}} \sum_{r=1}^{89} \frac{\sin (r+1)^{\circ} \cos r^{\circ}}{\sin r^{\circ} \cdot \sin (r+1)^{\circ}} \\ & \Rightarrow \frac{1}{\sin 1^{\circ}} \sum_{r=1^{\circ}}^{89} \cot r^{\circ}-\cot (r+1)^{\circ} \\ & \Rightarrow \frac{1}{\sin 1^{\circ}}\left[\cot 1^{\circ}-\cot 90^{\circ}\right]=\frac{\cot 1^{\circ}}{\sin 1^{\circ}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \cos ^3 \frac{\pi}{8} \cdot \cos \frac{3 \pi}{8}+\sin ^3 \frac{\pi}{8} \cdot \sin \frac{3 \pi}{8} \\ & \text { Using } \frac{3 \pi}{8}=\frac{\pi}{2}-\frac{\pi}{8} \\ & \Rightarrow \cos \frac{3 \pi}{8}=\sin \frac{\pi}{8} \text { and } \sin \frac{3 \pi}{8}=\cos \frac{\pi}{8} \\ & \Rightarrow \cos ^3 \frac{\pi}{8} \cdot \sin \frac{\pi}{8}+\sin ^3 \frac{\pi}{8} \cdot \cos \frac{\pi}{8} \\ & \quad=\sin \frac{\pi}{8} \cdot \cos \frac{\pi}{8}\left(\cos ^2 \frac{\pi}{8}+\sin ^2 \frac{\pi}{8}\right) \\ & \quad=\frac{1}{2} 2 \sin \frac{\pi}{8} \cos \frac{\pi}{8} \\ & \quad=\frac{1}{2} \sin \frac{\pi}{4}=\frac{1}{2 \sqrt{2}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, } A+B+C=\frac{\pi}{4}\\ &\begin{aligned} & \therefore 2 A+2 B+2 C=\frac{\pi}{2} \\ & \text { Now, } \sin 4 A+\sin 4 B+\sin 4 C \\ & 2 \sin (2 A+2 B) \cdot \cos (2 A-2 B)+2 \sin 2 C \\ & =2 \sin \left(\frac{\pi}{2}-2 C\right) \cdot \cos 2 C +2 \sin 2 C \cdot \cos 2 C \\ & =2 \cos 2 C \cdot \cos (2 A-2 B)+2 \sin 2 C \cdot \cos 2 C \\ & =2 \cos 2 C\{\cos (2 A-2 B)+\sin 2 C\} \\ & =2 \cos 2 C[\cos (2 A-2 B)+\cos (2 A+2 B)] \\ & =2 \cos 2 C[2 \cos 2 A \cdot \cos 2 B] \\ & =4 \cos 2 A \cdot \cos 2 B \cdot \cos 2 C \end{aligned} \end{aligned} $

Given, $630^{\circ}<\theta<810^{\circ}$ and

$ \tan \theta=\frac{-7}{24} $

$\therefore \theta$ lies IV quadrant

$ \begin{aligned} \therefore \quad \cos \theta & =\frac{24}{25} \quad 315^{\circ}<\frac{\theta}{2}<405^{\circ} \\ 2 \cos ^2 \frac{\theta}{2}-1 & =\frac{24}{25} \quad 157 \frac{1}{2}<\frac{\theta}{4}<2.02 \frac{1}{2} \\ 2 \cos ^2 \frac{\theta}{2} & =\frac{49}{25} \quad \frac{\theta}{4} \text { lies III quadrant } \\ \Rightarrow \quad \cos \frac{\theta}{2} & =\frac{7}{5 \sqrt{2}} \Rightarrow 2 \cos ^2 \frac{\theta}{4}-1=\frac{7}{5 \sqrt{2}} \\ \Rightarrow 2 \cos ^2 \frac{\theta}{4} & =1+\frac{7}{5 \sqrt{2}}=\frac{7+5 \sqrt{2}}{5 \sqrt{2}} \\ \Rightarrow \quad \cos \frac{\theta}{4} & =-\sqrt{\frac{7+5 \sqrt{2}}{10 \sqrt{2}}} \end{aligned} $

We have,

$ 2 \cos \theta+\sin \theta=1 \quad \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $

Squaring both sides, we get

$ \begin{aligned} & \Rightarrow \quad(2 \cos \theta)^2=(1-\sin \theta)^2 \\ & \Rightarrow \quad 4 \cos ^2 \theta=1+\sin ^2 \theta-2 \sin \theta \\ & \Rightarrow \quad 5 \sin ^2 \theta-2 \sin \theta-3=0 \\ & \Rightarrow \quad(\sin \theta-1)(5 \sin \theta+3)=0 \\ & \Rightarrow \quad \sin \theta=1 \text { and }-\frac{3}{5} \\ & \Rightarrow \quad \cos \theta=0 \text { and }+\frac{4}{5} \\ & \because \quad 7 \cos \theta+6 \sin \theta=k, \text { when } \sin \theta=1 \\ & \Rightarrow \quad 7(0)+6(1)=k \Rightarrow k=6 \end{aligned} $

$ \begin{aligned} &\text { When, } \sin \theta=-\frac{3}{5} \text { and } \cos \theta=\frac{4}{5}\\ &k=7\left(\frac{4}{5}\right)-6\left(\frac{3}{5}\right)=\frac{28-18}{5}=\frac{10}{5}=2 \end{aligned} $

$ \begin{aligned} &\sum_{k=0}^{12} \frac{1}{\sin \left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right) \cdot \sin \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)}\\ &\text { Multiplying } N^r \text { and } D^r \text { by } \sin \pi / 6 \text {, }\\ &\begin{aligned} & \frac{1}{\sin \frac{\pi}{6}} \sum_{k=0}^{12} \frac{\sin \left\{\left((k+1) \frac{\pi}{6}+\frac{4}{\pi}\right)-\left(\frac{k \pi}{6}+\frac{\pi}{4}\right)\right\}}{2 \sum_{k=0}^{12}\left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right) \cdot \sin \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)} \\ & =2\left[\cot \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)-\cot \left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right)\right\} \\ & \quad+\left\{\cot \left(\frac{\pi}{6}+\frac{\pi}{4}\right)-\cot \left(\frac{\pi}{6}+\frac{\pi}{4}\right)\right\} \\ & \quad+\ldots+\left\{\cot \left(\frac{12 \pi}{6}+\frac{\pi}{4}\right)-\cot \left(\frac{13 \pi}{6}+\frac{\pi}{4}\right)\right\} \\ & =2\left[\cot \frac{\pi}{4}+\cot \left(\frac{\pi}{6}+\frac{\pi}{4}\right)\right] \\ & =2\left[1-\frac{\sqrt{3}-1}{\sqrt{3}+1}\right]=2\left(\frac{2}{\sqrt{3}+1}\right)=\frac{4(\sqrt{3}-1)}{2} \\ & =2(\sqrt{3}-1) \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given } \cos \alpha=\sec h \beta\\ &\begin{aligned} & \text { or } \quad \cos h \beta=\sec \alpha \\ & \Rightarrow \quad \beta=\cos h^{-1}(\sec \alpha) \\ & \Rightarrow \quad \beta=\log \left(\sec \alpha+\sqrt{\sec ^2 \alpha-1}\right) \\ & {\left[\because \cos h^{-1} x=\log \left(x+\sqrt{x^2-1}\right)\right]} \\ & \beta=\log (\sec \alpha+\tan \alpha) \end{aligned} \end{aligned} $

Given that $(\sin \theta - \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 5$ and $\theta$ is in the third quadrant, we need to find $(\sin \theta + \cos \theta)^3$.

First, simplify the given expression:

$ (\sin \theta - \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 5 $

By expanding, we have:

$ \sin^2 \theta + \csc^2 \theta - 2 + \cos^2 \theta + \sec^2 \theta + 2 = 5 $

This simplifies to:

$ 1 + 1 + \cot^2 \theta + 1 + \tan^2 \theta = 5 $

Thus:

$ \tan^2 \theta + \cot^2 \theta = 2 $

This implies:

$ (\tan \theta - \cot \theta)^2 = 0 $

So, $\tan \theta = \cot \theta$, leading to $\tan^2 \theta = 1$. Therefore, $\tan \theta = \pm 1$.

Since $\theta$ lies in the third quadrant, where both sine and cosine are negative, we have $\tan \theta = 1$.

Thus, $\theta = \frac{5\pi}{4}$.

Now, find $(\sin \theta + \cos \theta)^3$:

$ \sin \theta = -\frac{1}{\sqrt{2}}, \quad \cos \theta = -\frac{1}{\sqrt{2}} $

Therefore:

$ (\sin \theta + \cos \theta) = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = -\sqrt{2} $

Thus:

$ (\sin \theta + \cos \theta)^3 = (-\sqrt{2})^3 = -2\sqrt{2} $

We have,

$ \begin{array}{l} \cos ^{2} B-\sin ^{2} A=\cos (A+B) \cdot \cos (A-B) \\\\ =\frac{\sqrt{3}+1}{2 \sqrt{2}} \cdot \frac{1}{2} \\\\ \therefore \cos (A+B) \cdot \cos (A-B)=\cos 15^{\circ} \cdot \cos 60^{\circ} \\\\ A+B=60^{\circ} \quad \ldots \text { (i) } \\\\ A-B=15^{\circ} \quad \ldots \text { (ii) } \\\\ A=\frac{75^{\circ}}{2} \text { and } B=\frac{45^{\circ}}{2} \end{array} $

Now, $ \cos ^{2} \frac{4}{3} \times \frac{45^{\circ}}{2}-\sin ^{2} \frac{4}{5} \times \frac{75^{\circ}}{2} $

$ =\cos ^{2} 30^{\circ}-\sin ^{2} 30^{\circ}=\cos 60^{\circ}=\frac{1}{2} $

$ 2 \sin ^{2} \theta=\cos ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8} +\cos ^{4} \frac{5 \pi}{8}+\sin ^{4} \frac{7 \pi}{8} $

$ =\cos ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8}+\cos ^{4}\left(\pi-\frac{3 \pi}{8}\right) +\sin ^{4}\left(\pi-\frac{\pi}{8}\right) $

$ =\cos ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8}+\cos ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{\pi}{8} $

$ 2 \sin ^{2} \theta=\left(\cos ^{4} \frac{\pi}{8}+\sin ^{4} \frac{\pi}{8}\right) +\left(\cos ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{3 \pi}{8}\right) $

$ \left[\because \cos ^{4} \theta+\sin ^{4} \theta=\left(\cos ^{2} \theta+\sin ^{2} \theta\right)^{2} -2 \sin ^{2} \theta \cos ^{2} \theta\right] $

$ =1-\frac{1}{2}(2 \sin \theta \cos \theta)^{2}=1-\frac{1}{2} \sin ^{2} 2 \theta $

$ \Rightarrow 2 \sin ^{2} \theta=\left(1-\frac{1}{2} \sin ^{2} \frac{\pi}{4}\right)+\left(1-\frac{1}{2} \sin ^{2} \frac{3 \pi}{4}\right) $

$ =1-\frac{1}{2} \times \frac{1}{2}+1-\frac{1}{2} \cdot \frac{1}{2}=2-\frac{1}{4}-\frac{1}{4} $

$ =2-\frac{1}{2}=\frac{3}{2} $

$ \Rightarrow 2 \sin ^{2} \theta=\frac{3}{2} $

$ \Rightarrow \sin ^{2} \theta=\frac{3}{4} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} $

$ \therefore \quad \theta=\frac{\pi}{3} $

$ 2 \tan ^{2} \theta-4 \sec \theta+3=0 \quad \text { [given] } $

$ \Rightarrow \quad 2\left(\sec ^{2} \theta-1\right)-4 \sec \theta+3=0 $

$ \Rightarrow \quad 2 \sec ^{2} \theta-4 \sec \theta+1=0 $

$ \Rightarrow \quad \sec \theta=\frac{4 \pm \sqrt{16-8}}{4}=\frac{4 \pm 2 \sqrt{2}}{4} $

$ \Rightarrow \quad \sec \theta=\frac{2(2 \pm \sqrt{2})}{4} $

$ \Rightarrow \quad 2 \sec \theta=2+\sqrt{2} $

and $ 2-\sqrt{2} $

$ \Rightarrow \quad 2 \sec \theta=2+\sqrt{2}(\because \operatorname{sech} \neq 2-\sqrt{2}) $