Problem-1

For$\quad$ $a>0, b>0$, and $c>0$ find

$\tan ^{-1} \sqrt{\frac{a(a+b+c)}{b c}}+\tan ^{-1} \sqrt{\frac{b(a+b+c)}{a c}} + \tan ^{-1} \sqrt{\frac{c(a+b+c)}{a b}}$

Solution

$\tan ^{-1}\left[a \sqrt{\frac{a+b+c}{a b c}}\right]+\tan ^{-1}\left[b \sqrt{\frac{a+b+c}{a b c}}\right] = \tan ^{-1} X+\tan ^{-1} Y$

where, $X=\left[a \sqrt{\frac{a+b+c}{a b c}}\right] , Y=\left[b \sqrt{\frac{a+b+c}{a b c}}\right]$

Observe, XY = $\frac{a + b + c}{c} > 1$

Solution

$\tan^{-1}\left[a\sqrt{\frac{a+b+c}{a b c}}\right]+\tan^{-1}\left[b\sqrt{\frac{a+b+c}{a b c}}\right]$

$\because X,Y>0\quad$ & $\quad XY>1$

$\Rightarrow \tan ^{-1} X + \tan ^{-1} Y = \pi+\tan ^{-1} \frac{X+Y}{1-X Y}$

$=\pi+\tan ^{-1} \frac{(a+b) \sqrt{\frac{a+b+c}{a b c}}}{1-\frac{(a+b+c)}{c}} =\pi+\tan ^{-1}\left[-c \sqrt{\frac{a+b+c}{a b}}\right]$

$=\pi-\tan^{-1}\left[c\sqrt{\frac{a+b+c}{a b c}}\right]$$\qquad\because\tan^{-1}(-x)=-\tan ^{-1} x$

Problem-2

For $\quad |x| \leqslant 1$ find the smallest and the greatest value of

$\left(\sin ^{-1} x\right)^4+\left(\cos ^{-1} x\right)^4$

Solution

Using the identity,

$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} .$

We rewrite the expression as follows:

$\left(\sin ^{-1} x\right)^4+\left(\cos ^{-1} x\right)^4=\left(\sin ^{-1} x\right)^4+\left(\frac{\pi}{2}-\sin ^{-1} x\right)^4$

Let, $\theta=\sin ^{-1} x, ~ ~$ where, $\theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Let, $f(\theta)=\theta^4+\left(\frac{\pi}{2}-\theta\right)^4, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Setting the derivative equal to zero, we get the point for extremum.

$\Rightarrow f^{\prime}(\theta)=4 \theta^3-4(\pi / 2-\theta)^3=0$

$\Rightarrow \theta^3-(\pi / 2-\theta)^3=0$

$\Rightarrow \theta^3=(\pi / 2-\theta)^3$

$\Rightarrow \theta=\frac{\pi}{2}-\theta\quad\Rightarrow \theta=\pi / 4$

To determine whether this point is a maximum or minimum, we need to take second derivative,

$f^{ \prime\prime}(\theta)=12 \theta^2+12(\frac{\pi}{2}-\theta)^2>0$

$\therefore\quad \theta=\frac{\pi}{4}$ will give a minimum value.

The minimum value of $f(\theta)$ is $(\frac{\pi}{4})^4$ + $(\frac{\pi}{2}-\frac{\pi}{4})^4= \frac {\pi^4}{ 128 }$

To find the maximum value we start with the first derivative of the given function,

$f^{\prime}(\theta)=4\left[\theta^3-(\pi / 2-\theta)^3\right]$

$\rightarrow$ For,$\frac{\pi}{4} \leqslant \theta \leqslant\frac{\pi}{2}$

$f^{\prime}(\theta) \geqslant 0 \Rightarrow f(\theta)$ is increasing in $[{\frac{\pi}{4}},{\frac{\pi}{2}}]$

$\rightarrow$ For $-\frac{\pi}{2} \leqslant \theta \leqslant \frac{\pi}{4}, \quad f^{\prime}(\theta) \leqslant 0$

$\Rightarrow f^{\prime}(\theta)$ is decreasing in $[{-\frac{\pi}{2}},{\frac{\pi}{4}}]$

We easily observe from the graph that the maximum value attains at end points.

$f(\theta)=\theta^4+\left(\frac{\pi}{2}-\theta\right)^4$

$\Rightarrow f\left(-\frac{\pi}{2}\right)=\frac{\pi^4}{16}+\pi^4$

$\Rightarrow f\left(\frac{\pi}{2}\right)=\frac{\pi^4}{16}$

The maximum value of

$\left(\sin ^{-1} x \right)^4+\left(\cos^{-1} x \right)^4$

$= \frac{\pi^4}{16}+\pi^4 = \frac{17 \pi^4}{18}$.

Problem-3

The number of solutions to the equation$\left|\sin ^{-1}(\sin x)\right|=\cos x, \text { where, }$

$x \in[-2 \pi, 2 \pi] \text { is }?$

Solution

$\sin ^{-1}(\sin x)=y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$\sin x=\sin y$

i) $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2} \quad \Rightarrow y=x$

ii) $\frac{\pi}{2} \leqslant x \leqslant \frac{3 \pi}{2},$ then, $\quad \pi-x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$\Rightarrow\sin (\pi-x)=\sin x=\sin y$

$\Rightarrow\pi-x=y\quad\Rightarrow \pi-x=y=\sin ^{-1}(\sin x)$

iii)$\quad\frac{3 \pi}{2} \leqslant x \leqslant \frac{5 \pi}{2}, \quad x-2 \pi \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

$\Rightarrow\sin (x-2 \pi)=\sin x=\sin y$

$\Rightarrow y=x-2 \pi$

$\rightarrow$ For $\frac{3\pi}{2}\leqslant x \leqslant \frac{5\pi}{2},\quad\sin^{-1}(\sin x)=x-2\pi$

$\rightarrow$ For $\frac{\pi}{2}\leqslant x \leqslant \frac{3\pi}{2},\quad\sin^{-1}(\sin x)=\pi-x$

$\rightarrow$ For $-\frac{\pi}{2}\leqslant x \leqslant \frac{\pi}{2},\quad\sin^{-1}(\sin x)=x$

$\rightarrow$ For $-\frac{3\pi}{2}\leqslant x \leqslant -\frac{\pi}{2},\quad\sin^{-1}(\sin x)=-(\pi+x)$

$\rightarrow$ For $-\frac{5\pi}{2}\leqslant x \leqslant -\frac{3\pi}{2},\quad\sin^{-1}(\sin x)=-(x+2\pi)$

Problem-4

Let, $\quad S = \left\{x \in(-\pi, \pi): x \neq 0, \pm \frac{\pi}{2}\right\}$ the sum of all distinct solutions of the equation

$\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)$ $=0$

in the set S is equal to ?

Solution

$\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$

$\because\quad\sin x \neq 0\quad$,$\cos x \neq 0$

$\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}+2\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)=0$

if $x \in$S then, $\quad\sin x \cos x \neq 0$

$\sqrt{3} \sin x+\cos x+2\left(\sin ^2 x-\cos ^2 x\right)=0$

$\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\cos 2 x$

$\cos x \cos \frac{\pi}{3}+\sin x \sin \frac{\pi}{3}=\cos 2 x$

$\cos \left(x-\frac{\pi}{3}\right)=\cos 2 x$

$\cos x=\cos y \Leftrightarrow x=2 n \pi \pm y$

$\cos \left(x-\frac{\pi}{3}\right)=\cos 2 x$

by using identity y = $x-\frac{\pi}{3}$

$2 x=2 n \pi \pm(x-\pi / 3)$

$2 x=2 n \pi+(x-\pi / 3)$

$x=2 n \pi-\pi / 3$

$n=0, x=-\pi / 3 \in(-\pi, \pi), n=1, x=2 \pi-\frac{\pi}{3} \notin(-\pi, \pi)$

$x=-\pi / 3$

$2 x=2 n \pi-(x-\pi / 3)$

$3 x=2 n \pi+\pi / 3$

$x=\frac{2 n \pi}{3}+\pi / 9$

$n=0, \quad x=\pi / 9 \in(-\pi, \pi)$

$n=1, \quad x=\frac{2 \pi}{3}+\pi / 9 \in(-\pi, \pi)$

$n=-1, \quad x=\frac{-2 \pi}{3}+\pi / 9 \in(-\pi, n)$

$x=-\frac{\pi}{3}$

$\cos 2 x =\cos (x-\pi / 3)=\cos (\pi / 3-x)$

$2 x =\operatorname{sn\pi } \pi \pm(x-\pi / 3)$

$\cos 2 x =\cos (\pi / 3-x)$

$2 x =2 n \pi \pm(\pi / 3-x)$