Trigonometric Ratios & Identities

$\sinh (x+y) \cosh (x-y)=\frac{1}{2} [2 \sin \mathrm{h}(x+y) \cosh (x-y)]$ .... (i)

$\because \sin \mathrm{h} A+\sin \mathrm{h} B=2 \sin \mathrm{h}\left(\frac{A+B}{2}\right) \cosh \left(\frac{A-B}{2}\right)$

Let $\frac{A+B}{2}=x+y$

$\begin{aligned} & \Rightarrow A+B=2 x+2 y \\ & \text { and } A-B=2 x-2 y \end{aligned}$

Adding, we obtain $2 A=4 x$

$A=2 x$

Subtracting, we obtain $2 B=4 y$

$\begin{aligned} B & =2 y \\ \therefore 2 \sinh (x+y) \cosh (x-y) & =\sinh 2 x+\sin \mathrm{h} 2 y \\ \Rightarrow \sinh (x+y) \cosh (x-y) & =\frac{1}{2}[\sin \mathrm{h} 2 x+\sin \mathrm{h} 2 y] \end{aligned}$

$\begin{aligned} \cos \left(\frac{\pi}{8}\right) & =\sqrt{\frac{1+\cos \frac{\pi}{4}}{2}} \\ & =\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}}=\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}\end{aligned}$

$\begin{aligned} & \cos \theta=-\sqrt{\frac{3}{2}} \\ & \sin \alpha=-\frac{3}{5} \\ & \text { As, }-1 \leq \cos \theta \leq 1 \end{aligned}$

In the question given, $\cos \theta=-\sqrt{\frac{3}{2}}$. So, this is a wrong question.

$\begin{aligned} & \tan \beta=\frac{\tan \alpha+\tan \gamma}{1+\tan \alpha \tan \gamma} \\ & \tan \beta=\frac{\sin \alpha \cos \gamma+\sin \gamma \cos \alpha}{\cos \alpha \cos \gamma+\sin \alpha \sin \gamma} \\ & \tan \beta=\frac{\sin (\alpha+\gamma)}{\cos (\alpha-\gamma)} \\ & \frac{\sin 2 \alpha+\sin 2 \gamma}{1+\sin 2 \alpha \sin 2 \gamma} \\ & =\frac{2 \sin (\alpha+\gamma) \cos (\alpha-\gamma)}{1+(2 \sin \alpha \cos \gamma)(2 \sin \gamma \cos \alpha)} \\ & =\frac{2 \sin (\alpha+\gamma) \cos (\alpha-\gamma)}{1+[\sin (\alpha+\gamma)+\sin (\alpha-\gamma)][\sin (\alpha+\gamma)-\sin (\alpha-\gamma)]}\end{aligned}$

$\begin{aligned} & =\frac{2 \sin (\alpha+\gamma) \cos (\alpha-\gamma)}{1+\sin ^2(\alpha+\gamma)-\sin ^2(\alpha-\gamma)} \\ & =\frac{2 \sin (\alpha+\gamma) \cos (\alpha-\gamma)}{\sin ^2(\alpha+\gamma)+\cos ^2(\alpha-\gamma)}=\frac{2 \tan \beta}{1+\tan ^2 \beta}=\sin 2 \beta\end{aligned}$

$\begin{aligned} & \cos \left(\alpha+\frac{\pi}{2}\right)=-\sin \alpha \\ & \cos \left(\beta+\frac{\pi}{2}\right)=-\sin \beta \\ & \cos \left(\gamma+\frac{\pi}{2}\right)=-\sin \gamma \\ & \because \quad A M \geq G M \\ & \frac{-(\sin \alpha+\sin \beta+\sin \gamma)}{3} \geq(-\sin \alpha \sin \beta \sin \gamma)^{1 / 3} \end{aligned}$

Equating holds, when $\alpha=\beta=\gamma$

$\begin{gathered} \alpha+\beta+\gamma=360 \gamma \\ \alpha=\beta=\gamma=120 \gamma \\ \Rightarrow A M=\frac{1}{3}\left[\left(-\frac{\sqrt{3}}{2}\right)+\left(-\frac{\sqrt{3}}{2}\right)+\left(-\frac{\sqrt{3}}{2}\right)\right] \\ =-\frac{\sqrt{3}}{2} \end{gathered}$

$3\sin A + 4\cos B = 6$ ...... (i)

and $4\sin B + 3\cos A = 1$ .... (ii)

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

$9{\sin ^2}A + 16{\cos ^2}B + 24\sin A\cos B = 36$ ..... (iii)

$16{\sin ^2}B + 9{\cos ^2}A + 24\sin B\cos A = 1$ .... (iv)

Adding Eqs. (iii) and (iv), we get

$\begin{aligned} & 9\left(\sin ^2 A+\cos ^2 A\right)+16\left(\sin ^2 B+\cos ^2 B\right) \\ & +24(\sin A \cos B+\cos A \sin B)=37 \\ & \Rightarrow \quad 25+24 \sin (A+B)=37 \\ & \sin (A+B)=1 / 2 \\ & \end{aligned}$

Let $\tan A+2 \cot 2 A$

$= \frac{\sin A}{\cos A}+\frac{2 \cos 2 A}{\sin 2 A}$

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

$= \frac{\sin ^2 A+1-2 \sin ^2 A}{\sin A \cos A}$

$= \frac{1-\sin ^2 A}{\sin A \cos A}=\frac{\cos ^2 A}{\sin A \cos A}=\cot A$

$\therefore \tan \alpha+2 \tan 2 \alpha+4 \tan 4 \alpha+8 \cot 8 \alpha \\ = \tan \alpha+2 \tan 2 \alpha+4(\tan 4 \alpha+2 \cot 2(4 \alpha))$

$= \tan \alpha+2 \tan 2 \alpha+4 \cot 4 \alpha$

$= \tan \alpha+2(\tan 2 \alpha+2 \cot 2(2 \alpha))$

$= \tan \alpha+2 \cot 2 \alpha=\cot \alpha$

$\begin{aligned} & \text { If } f(x)=\frac{\cot x}{1+\cot x} \text { and } \alpha+\beta=\frac{5 \pi}{4} \\ & \Rightarrow \cot (\alpha+\beta)=\cot \frac{5 \pi}{4}=1 \Rightarrow \frac{\cot \alpha \cot \beta-1}{\cot \beta+\cot \alpha}=1 \\ & \Rightarrow \cot \alpha \cot \beta=\cot \beta+\cot \alpha+1\quad ....(\mathrm{i}) \\ & \text { Now, } f(\alpha) f(\beta)=\frac{\cot \alpha}{1+\cot \alpha} \times \frac{\cot \beta}{1+\cot \beta} \\ & =\frac{\cot \alpha \cot \beta}{1+\cot \alpha+\cot \beta+\cot \alpha \cot \beta}=\frac{\cot \alpha \cot \beta}{2 \cot \alpha \cot \beta}=\frac{1}{2}\end{aligned}$

In $\triangle A B C$,

$\begin{aligned} & A B=c, B C=a, A C=b \\ & \because \frac{a+b+c}{B C+A B}+\frac{a+b+c}{A C+A B}=3 \\ & \Rightarrow \frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3 \\ & \Rightarrow 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3 \\ & \Rightarrow \quad \frac{b}{a+c}+\frac{a}{b+c}=1 \\ & \Rightarrow \quad \frac{a^2+b^2-c^2}{}=a b \\ & \Rightarrow \quad \frac{a^2+b^2-c^2}{2 a b}=\frac{1}{2} \Rightarrow \cos C=\frac{1}{2} \\ & \Rightarrow \quad C=60 \Upsilon \end{aligned}$

Now, $\tan \frac{C}{8} =\tan \frac{60 \Upsilon}{8}$

$=\tan \left(7 \frac{1}{2}\right)^{\Upsilon}=(\sqrt{3}-\sqrt{2})(\sqrt{2}-1)$

$=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

To find the mean of the values given, $\sin^2 10Y, \sin^2 20Y, \sin^2 30Y, \ldots, \sin^2 90Y$, we first need to recognize a pattern or symmetry that can simplify our calculations.

Firstly, we note that the sine function has a property where $\sin^2$ of complementary angles add to 1. This is given by the identity:

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

When dealing with complementary angles, we can write this as:

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

So, if we pair each element with its complement to $90^\circ$, we get:

$\sin^2 10Y = \cos^2 80Y$

$\sin^2 20Y = \cos^2 70Y$

$\sin^2 30Y = \cos^2 60Y$

$\sin^2 40Y = \cos^2 50Y$

$\sin^2 50Y = \cos^2 40Y$

$\sin^2 60Y = \cos^2 30Y$

$\sin^2 70Y = \cos^2 20Y$

$\sin^2 80Y = \cos^2 10Y$

$\sin^2 90Y = 1$

Essentially, the sum of each pair $\sin^2 \theta + \cos^2 \theta = 1$. Therefore, pairing these up (excluding $\sin^2 90Y$ as it stands alone giving 1), we see that the total number of pairs is 4 (since we started from 10 up to 80, which is 8 elements). Each pair sums to 1, so we have:

$\sin^2 10Y + \sin^2 80Y = 1$

$\sin^2 20Y + \sin^2 70Y = 1$

$\sin^2 30Y + \sin^2 60Y = 1$

$\sin^2 40Y + \sin^2 50Y = 1$

And $\sin^2 90Y = 1$

Thus, the total sum of the 9 terms is:

$4 \times 1 + 1 = 5$

Since there are 9 total terms, the mean is:

$\frac{\text{Total sum}}{\text{Number of terms}} = \frac{5}{9}$

Therefore, the correct answer is Option A: $\frac{5}{9}$

Let coordinate of $P$ original system be $(x, y)$.

Then, new coordinate after rotating axes through $135\Upsilon$ is $(4,-3)$

$\begin{aligned} & \therefore \quad 4=x \cos 135 x+y \sin 135\Upsilon \\ & \text { and }-3=-x \sin 135 \gamma+y \cos 135\Upsilon\\ & \Rightarrow \quad 4=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}} \\ & \text { and } \quad-3=\frac{-x}{\sqrt{2}}-\frac{y}{\sqrt{2}} \\ \end{aligned}$

On adding, $\quad 1=\frac{-2 x}{\sqrt{2}} \Rightarrow x=-\frac{1}{\sqrt{2}}$

On subtracting, $7=\frac{2 y}{\sqrt{2}} \Rightarrow y=\frac{7}{\sqrt{2}}$

Coordinate of $P$ in original system $\left(\frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$.

Given, $f(x)=\sin x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Range of $\sin x=[-1,1]$

For $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \sin x$ is increasing function

It is maximum at $x=\frac{\pi}{2}$

$\operatorname{Max} f(x)=\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}=1$

$\begin{aligned} & \because \tan (A+B+C) \\ & \quad=\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A} \end{aligned}$

If $A+B+C=90 \Upsilon$, then

$1-\tan A \tan B-\tan B \tan C-\tan C \tan A=0$

$\text { or } \tan A \tan B+\tan B \tan C+\tan C \tan A=1$

Here, $A=2 \alpha, B=30 \Upsilon-\alpha, C=60 \Upsilon-\alpha$, then

$\begin{aligned} & A+B+C=2 \alpha+30 Y-\alpha+60 Y-\alpha=90^{\circ} \\ & \therefore \tan 2 \alpha \tan (30 Y-\alpha)+\tan 2 \alpha \cdot \tan (60 Y-\alpha) \\ & +\tan (30 Y-\alpha) \tan (60 Y-\alpha)=1 \end{aligned}$

$\sin \alpha - \cos \alpha = m$ and $\sin 2\alpha = n - {m^2}$

$\therefore$ ${(\sin \alpha - \cos \alpha )^2} = {m^2}$

${\sin ^2}\alpha + {\cos ^2}\alpha - 2\sin \alpha \cos \alpha = {m^2}$

or $1 - \sin 2\alpha = {m^2}$

or $\sin 2\alpha = 1 - {m^2}$

But $\sin 2\alpha = n - {m^2}$ (given)

$ \Rightarrow n = 1$

$\begin{aligned} & \text { } \because \tan \theta=\sinh u \\ & \Rightarrow \quad \tan \theta=\sqrt{\cosh ^2 u-1} \quad\left[\cosh ^2 x-\sinh ^2 x=1\right] \\ & \Rightarrow \cosh u=\sqrt{1+\tan ^2 \theta} \Rightarrow \cosh u=\sec \theta\end{aligned}$

Given,

$ \begin{aligned} & A=\left[\begin{array}{ccc} \cos ^2 37 & \cos ^2 53 & \cot 135 \\ \sin ^2 76 & \sin 270 & \sin ^2 14 \\ \cos 180 & \cos ^2 28 & \cos ^2 62 \end{array}\right] \\ & \therefore A=\left[\begin{array}{ccc} \cos ^2 37 & \cos ^2 53 & -1 \\ \sin ^2 76 & -1 & \sin ^2 14 \\ -1 & \cos ^2 28 & \cos ^2 62 \end{array}\right] \\ & \therefore|A|=\left|\begin{array}{ccc} \cos ^2 37 & \cos ^2 53 & -1 \\ \sin ^2 76 & -1 & \sin ^2 14 \\ -1 & \cos ^2 28 & \cos ^2 62 \end{array}\right| \end{aligned} $

Use $C_1 \rightarrow C_1+C_2$ and $C_2 \longrightarrow C_2+C_3$

$ |A|=\left\lvert\, \begin{array}{cc} \cos ^2 37+\cos ^2 53 & \cos ^2 53-1 \\ \sin ^2 76-1 & -1+\sin ^2 14 \\ -1+\cos ^2 28 & \cos ^2 28+\cos ^2 62 \end{array}\right. $

$\left. \matrix{ - 1\,\,\,\,\,\,\,\,\,\,\, \hfill \cr {\sin ^2}14\,\,\, \hfill \cr {\cos ^2}62\, \hfill \cr} \right|$$ \begin{aligned} & |A|=\left|\begin{array}{ccc} 1 & -\sin ^2 53 & -1 \\ -\cos ^2 76 & -\cos ^2 14 & \sin ^2 14 \\ -\sin ^2 28 & 1 & \cos ^2 62 \end{array}\right| \\ & |A|=-\left|\begin{array}{ccc} 1 & -\sin ^2 53 & +1 \\ -\sin ^2 14 & -\cos ^2 14 & -\sin ^2 14 \\ -\cos ^2 62 & 1 & -\cos ^2 62 \end{array}\right| \\ & \Rightarrow|A|=0 \\ & \text { Now 3-|A|=3 } \\ & \therefore \quad \text { (A) → (iii) } \end{aligned} $

(B) Given $\frac{\cos (6 x-4)+\sec (3-4 x)}{\cot (5 x+3)+\sin (3 x+4)}$

∴ Period of $\cos (6 x-4)=\frac{2 \pi}{6}=\frac{\pi}{3}$

Period of $\sec (3-4 x)=\frac{2 \pi}{4}=\frac{\pi}{2}$

Period of $\cot (5 x+3)=\frac{\pi}{5}$

Period of $\sin (3 x+4)=\frac{2 \pi}{3}$

Now required period = lim of $\left(\frac{\pi}{3}, \frac{\pi}{2}, \frac{\pi}{5}, \frac{2 \pi}{3}\right)=2 \pi$

Now, given $\frac{2 k \pi}{5}=2 \pi$

$ \Rightarrow k=5 $

(B) $\rightarrow(\mathrm{V})$

(C) given

$ \begin{aligned} & y=\cos ^2\left(\frac{\pi}{4}-x\right)+(\sin x-\cos x)^2 \\ & \therefore y=\frac{1}{2}(\cos x+\sin x)^2+(\sin x-\cos x)^2 \\ & \quad y=\frac{1}{2}(1+\sin 2 x)+(1-\sin 2 x) \\ & \Rightarrow \quad y=\frac{3}{2}-\frac{\sin 2 x}{2} \end{aligned} $

$\therefore y_{\text {max }}$ when $\sin 2 x=-1$

$ \therefore \quad y_{\max }=\frac{3}{2}+\frac{1}{2}=2 $

(C) → (ii)

(D) It $x+y+z=0^{\circ}$

$ \begin{array}{r} \Rightarrow \sin 2 x+\sin 2 y+\sin 2 z \\ =-4 \sin x \sin y \sin z \end{array} $

Now, $\frac{\sin 2 x+\sin 2 y+\sin 2 z}{\sin (-x) \sin (-y) \sin (-z)}$

$ \Rightarrow \quad \frac{-4 \sin x \sin y \sin z}{-\sin x \sin y \sin z}=4 $

(D) → (iv)

$ \begin{aligned} & \text { Given, } \cos (3 x+5)+7 \\ & \therefore \quad \text { Period }=\frac{2 \pi}{|3|} \\ & \left\{\because \text { Period of } \cos (a x+b)+c=\frac{2 \pi}{|a|}\right\} \\ & \text { Period }=\frac{2 \pi}{3} \end{aligned} $

Given, $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$

$ \therefore \quad \frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{2}{1} $

Now, using componendo and dividendo,

$ \begin{aligned} & \Rightarrow \quad \frac{\cos \left(\frac{\alpha-\beta}{2}\right)+\cos \left(\frac{\alpha+\beta}{2}\right)}{\cos \left(\frac{\alpha-\beta}{2}\right)-\cos \left(\frac{\alpha+\beta}{2}\right)}=\frac{2+1}{2-1} \\ & \Rightarrow \quad \frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}}=3 \\ & \Rightarrow \quad \tan \frac{\alpha}{2} \tan \frac{\beta}{2}=\frac{1}{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \quad \begin{aligned} \cos x-\sin x & =\sqrt{a} \sin x \\ \Rightarrow\,\,\,\,\,\,\,\,\,\,\,\cos x & =\sin x(\sqrt{a}+1) \end{aligned} \\ & \cos x=\frac{\sin x(\sqrt{a}+1)(\sqrt{a}-1)}{(\sqrt{a}-1)} \\ &(\sqrt{a}-1) \cos x=a \sin x-\sin x \\ & \sqrt{a} \cos x=a \sin x+\cos x-\sin x \\ & \therefore \quad a \sin x+\cos x-\sin x=\sqrt{a} \cos x \end{aligned} \end{aligned} $

A. Period of $\sin ^2 x$ is $\pi$

$ \begin{aligned} f(x) & =\sin ^2 x \\ f(\pi+x) & =\sin ^2(\pi+x) \\ & =(-\sin x)^2 \\ & =\sin ^2 x=f(x) \end{aligned} $

B. $\frac{\pi}{3}(\sqrt{3} \cos 3 x+\sin 3 x)$

$ \begin{aligned} \text { Maximum value } & =\frac{\pi}{3} \sqrt{(\sqrt{3})^2+(1)^2} \\ & =\frac{\pi}{3} \times 2=\frac{2 \pi}{3} \end{aligned} $

C. We have, $\sin \frac{x}{3}+\cos \frac{x}{2}$

Period of $\sin x$ is $6 \pi$ and period of $\cos \frac{x}{2}$ is $4 \pi$.

∴ Period of $\sin \frac{x}{3}+\cos \frac{x}{2}$ is LCM of $6 \pi$ and $4 \pi$ i.e., $12 \pi$.

D. We have, $y=|\sin x|$ and $y=1$

$ \begin{array}{lrl} \Rightarrow & 1 & =|\sin x| \\ \Rightarrow & \sin x & = \pm 1 \\ \Rightarrow & & x = \frac{\pi}{2} \text { in }[0, \pi] \end{array} $

∴ Intersection part is $\frac{\pi}{2}$.

$ \begin{aligned} & \text { We have, } \cot \frac{A}{2}=\sqrt{\frac{1+a}{1-a}} \cot \frac{\theta}{2} \\ & \Rightarrow \quad \cot ^2 \frac{A}{2}=\left(\frac{1+a}{1-a}\right) \cot ^2 \frac{\theta}{2} \\ & \Rightarrow \quad\left(\frac{\cos ^2 \frac{A}{2}}{\sin ^2 \frac{A}{2}}\right)\left(\frac{1-a}{1+a}\right)=\frac{\cos ^2 \theta / 2}{\sin ^2 \theta / 2} \\ & \Rightarrow \quad \frac{(1+\cos A)(1-a)}{(1-\cos A)(1+a)}=\frac{1+\cos \theta}{1-\cos \theta} \\ & \Rightarrow \frac{1-a-a \cos A+\cos A}{1+a-a \cos A-\cos A}=\frac{1+\cos \theta}{1-\cos \theta} \end{aligned} $

Apply componendo and dividendo,

$ \begin{array}{rlrl} \Rightarrow & & \frac{2(1-a \cos A)}{2(\cos A-a)} & =\frac{2}{2 \cos \theta} \\ \Rightarrow & \cos \theta & =\frac{\cos A-a}{1-a \cos A} \end{array} $

$ \begin{aligned} &\text { We have, } \sin \theta \cosh \alpha=\tan x\\ &\begin{gathered} \cos \theta \sinh \alpha=\sec x \\ \cos ^2 \theta \sin ^2 \mathrm{~h} \alpha-\sin ^2 \theta \cos ^2 \mathrm{~h} \alpha \\ =\sec ^2 x-\tan ^2 x \\ \cos ^2 \theta \sin ^2 \mathrm{~h} \alpha-\sin ^2 \theta \cos ^2 \mathrm{~h} \alpha=1 \\ \cos ^2 \theta \sin ^2 \mathrm{~h} \alpha-\cos ^2 \mathrm{~h} \alpha \\ +\cos ^2 \theta \cos ^2 \mathrm{~h} \alpha=1 \\ \cos ^2 \theta\left(\sin ^2 \mathrm{~h} \alpha+\cos ^2 \mathrm{~h} \alpha\right)-\cos ^2 \mathrm{~h} \alpha=1 \\ \cos ^2 \theta \cosh 2 \alpha-\cos ^2 \mathrm{~h} \alpha=1 \\ \cos ^2 \theta \cosh 2 \alpha-\left(\frac{\cosh 2 \alpha+1}{2}\right)=1 \\ 2 \cos ^2 \theta \cosh 2 \alpha-\cosh 2 \alpha-1=2 \\ \cosh 2 \alpha\left(2 \cos ^2 \theta-1\right)=3 \\ \Rightarrow \quad \cos 2 \theta \cosh 2 \alpha=3 \end{gathered} \end{aligned} $

$ \begin{aligned} &\text { Maximum value of } 3 \cos \theta-4 \sin \theta\\ &\begin{array}{ll} & a=\sqrt{3^2+(-4)^2}=5 \\ \therefore & \alpha=5 \sin ^2 \theta \cos ^3 \theta \\ & \beta=5 \sin ^3 \theta \cos ^2 \theta \end{array} \end{aligned} $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} & \alpha^2+\beta^2=5^2\left(\sin ^4 \theta \cos ^6 \theta+\sin ^6 \theta \cos ^4 \theta\right) \\ & =25\left(\cos ^2 \theta+\sin ^2 \theta\right) \sin ^4 \theta \cos ^4 \theta \\ & =25 \sin ^4 \theta \cos ^4 \theta \\ & \therefore \quad\left(\alpha^2+\beta^2\right)^5=\left(25 \sin ^4 \theta \cos ^4 \theta\right)^5 \\ & \Rightarrow\left(\alpha^2+\beta^2\right)^{5 / 2}=\left(5 \sin ^2 \theta \cos ^2 \theta\right)^5 \\ & \quad \alpha \beta=25 \sin ^5 \theta \cos ^5 \theta \\ & \Rightarrow \quad(\alpha \beta)^2=\left(5^2 \sin ^5 \theta \cos ^5 \theta\right)^2 \\ & \therefore \quad \sqrt{\frac{\left(\alpha^2+\beta^2\right)^5}{(\alpha \beta)^4}}=\frac{\left(5 \sin ^2 \theta \cos ^2 \theta\right)^5}{\left(5^2 \sin ^5 \theta \cos ^5 \theta\right)^2} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad=\frac{5^5 \sin ^{10} \theta \cos ^{10} \theta}{5^4 \sin ^{10} \theta \cos ^{10} \theta}=5 \end{aligned} \end{aligned} $

Given,

$\sin A=\frac{11}{61} A$ lies in 2nd quadrant

$\cos B=\frac{-7}{25} B$ lies in 3rd quadrant

$\cos A=\frac{-60}{61}$ and $\sin B=\frac{-24}{25}$

$ \begin{array}{r} \because \sin (A-B)=\sin A \cos B-\cos A \sin B \\ \quad=\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)-\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right) \end{array} $

$ \begin{aligned} & \sin (A-B)<0 \\ & \cos (A-B)=\cos A \cos B+\sin A \sin B \\ & =\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)+\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right) \end{aligned} $

$ \cos (A-B)>0 $

$\because A-B$ lie in 4th quadrant

$ \begin{aligned} & \sin (A+B)=\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)+\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right) \\ & \sin (A+B)>0 \\ & \cos (A+B)=\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)-\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right) \\ & \cos (A+B)>0 \end{aligned} $

$\therefore(A+B)$ lies in Ist quadrant

$ \begin{aligned} & \text { Given, } \\ & \cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x \\ & =\cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x \\ & \cos 2 x\left[\cos \left(\frac{\pi}{4}-x\right)+\cos \left(\frac{\pi}{4}+x\right)\right] \\ & =\sec x \sin 2 x(\cos x-\sin x) \\ & \cos 2 x\left(2 \sin \frac{\pi}{4} \sin x\right) \\ & =\sec x \sin 2 x(\cos x-\sin x) \\ & \begin{array}{c} \frac{2}{\sqrt{2}}\left(\cos ^2 x-\sin ^2 x\right) \sin x \\ =\sec x(2 \sin x \cos x)(\cos x-\sin x) \\ \cos x+\sin x=\sqrt{2} \end{array} \\ & \begin{array}{r} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because\,\, sin \,\,x=\cos x=\frac{1}{\sqrt{2}} \\ \sec x=\sqrt{2} \end{array} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8} \\ & +\cos ^4 \frac{7 \pi}{8}-\sin ^4 \frac{7 \pi}{8} \\ & =\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8} \\ & +\sin ^4\left(\pi-\frac{3 \pi}{8}\right)+\cos ^4 \frac{7 \pi}{8}-\sin ^4\left(\pi-\frac{\pi}{8}\right) \\ & =\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{3 \pi}{8} \\ & +\cos ^4 \frac{7 \pi}{8}-\sin ^4 \frac{\pi}{8}=\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{7 \pi}{8} \\ & =\cos ^4 \frac{3 \pi}{8}+\cos ^4\left(\pi-\frac{\pi}{8}\right) \\ & =\cos ^4 \frac{3 \pi}{8}+\left(-\cos \frac{\pi}{8}\right)^4 \\ & =\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{\pi}{8} \end{aligned} \end{aligned} $

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

$ \begin{aligned} & \text { Reason In } \triangle P Q R, \\ & P+Q+R=180^{\circ} \\ & \Rightarrow \quad \frac{P}{2}+\frac{Q}{2}+\frac{R}{2}=90^{\circ} \\ & \Rightarrow \quad \frac{P}{2}+\frac{Q}{2}=90^{\circ}-\frac{R}{2} \\ & \Rightarrow \quad \tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\tan \left(90-\frac{R}{2}\right) \\ & \Rightarrow \quad \tan \frac{P}{2}+\tan \frac{Q}{2} \\ & \Rightarrow \quad \frac{P}{1-\tan \frac{P}{2} \tan \frac{Q}{2}}=\cot \frac{R}{2} \\ & \Rightarrow\left(\tan \frac{P}{2}+\tan \frac{Q}{2}\right) \tan \frac{R}{2}=1-\tan \frac{P}{2} \tan \frac{Q}{2} \\ & \Rightarrow \tan \frac{P}{2} \tan \frac{Q}{2}+\tan \frac{Q}{2} \tan \frac{R}{2} +\tan \frac{R}{2} \tan \frac{P}{2}=1 \end{aligned} $

So, reason is true.

Assertion

We have,

$ \begin{aligned} & A=15^{\circ}, B=17^{\circ}, C=13^{\circ} \\ \Rightarrow & A+B+C=45^{\circ} \\ \Rightarrow & 2 A+2 B+2 C=90^{\circ} \\ \therefore & \frac{P}{2}=2 A, \frac{Q}{2}=2 B, \frac{R}{2}=2 C \\ \therefore & \tan 2 A \tan 2 B+\tan 2 B \tan 2 C +\tan 2 C \tan 2 A=1 \\ \Rightarrow & \frac{1}{\cot 2 A \cot 2 B}+\frac{1}{\cot 2 B \cot 2 C} +\frac{1}{\cot 2 C \cot 2 A}=1 \\ \Rightarrow & \frac{\cot 2 C+\cot 2 A+\cot 2 B}{\cot 2 A \cot 2 B \cot 2 C}=1 \\ \Rightarrow & \cot 2 A+\cot 2 B+\cot 2 C \\ = & \cot 2 A \cot 2 B \cot 2 C \\ \therefore & \text { Assertion is true. } \end{aligned} $