Problem:1

Prove that,

$\cos \left[\tan ^{-1}\left[\sin \left(\cot ^{-1} x\right)\right]\right]=\sqrt{\frac{x^2+1}{x^2+2}}$ .

Solution:

Let, $\cot ^{-1} x=\theta \in(0, \pi) \Rightarrow$ $x=\cot (\theta)$ , $\tan \theta=\frac{1}{x}$

$\sin \theta =\sqrt{\sin ^2 \theta} \quad \because \theta \in(0,\pi) \Rightarrow \sin \theta > 0$

$=\sqrt{1-\cos ^2 \theta}$

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

$=\sqrt{1-\frac{1}{1+\tan ^2 \theta}}$

$=\sqrt{\frac{\tan ^2 \theta}{1+\tan ^2 \theta}}$

$\sin \theta=\sqrt{\frac{\frac{1}{x^2}}{1+\frac{1}{x^2}}}=\sqrt{\frac{1}{1+x^2}}$ $\Rightarrow \sin (\cot^{-1})=\sqrt{\frac{1}{1+x^2}}$

Now, substituting in the expression we get,

$\cos \left[\left(\tan ^{-1} \frac{1}{\sqrt{1+x^2}}\right)\right]$

Let, $\tan ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)=\phi \in\left(0, \frac{\pi}{2}\right) \because \theta > 0$

$\tan \phi=\frac{1}{\sqrt{1+x^2}}$ , again substitute the expression of $\tan^{-1}(\frac{1}{1+x^{2}})$ we get,

$\cos \phi=\sqrt{\cos ^2 \phi}=\frac{1}{\sqrt{\sec ^2 \phi}}=\frac{1}{\sqrt{1+\tan ^2 \phi}} \cdots (a)$

We know,

$\tan \phi=\frac{1}{\sqrt{1+x^2}}$ on substitution in $\cdots$ (a) we get,

$\cos \phi=\frac{1}{\sqrt{1+\tan ^2 \phi}}$ $=\frac{1}{\sqrt{1+\frac{1}{1+x^2}}}$

$=\sqrt{\frac{1+x^2}{2+x^2}}$

Problem:2

If $f:[0,4 \pi] \rightarrow[0, \pi]$ be defined by $f(\theta)=\cos ^{-1}(\cos \theta)$. Then the number of points $\theta \in[0,4 \pi]$ satisfying the equation $f(\theta)=\frac{10-\theta}{10}$ is ____?

Solution:

$\theta \in[0,4 \pi]$ such that

$\cos ^{-1}(\cos \theta)=\frac{10-\theta}{10}$

${\theta \in[0,4 \pi]}$

Divide the region $[0,4\pi]$ into 4 regions,

$\quad \theta \in(\pi, 2 \pi] \Leftrightarrow(2\pi-\theta) \in[0, \pi] ⊆$ Range set of $\cos ^{-1}$

$\cos ^{-1}(\cos \theta)≠\theta$

{$\because \cos ^{-1}(\cos \theta) \in [0,\pi]$ and $\theta \in[\pi, 2\pi]$ }

$\cos ^{-1}(\cos \theta)=x , x\in[0, \pi]$

$\cos \theta=\cos x$

$\cos (\underbrace{2 \pi-\theta}_\phi)=(\cos \theta)$

$\cos \phi=\cos \theta$

$\phi \in$ Range set of $\cos^{-1}$

$\phi=\cos ^{-1}(\cos \theta)=2 \pi-\theta$

If $\theta \in(2 \pi, 3 \pi]$

$\theta-2 \pi \in{(0, \pi]} ⊆$ Range set of $\cos ^{-1}$

$\cos (\theta-2 \pi)=\cos \theta$

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

If $\theta \in(3 \pi, 4 \pi]$

$(4 \pi-\theta) \in[0, \pi) ⊆$ Range set of $\cos ^{-1}$

$\cos (4 \pi-\theta)=\cos \theta$

$4\pi-\theta=\cos ^{-1}(\cos \theta)$

Problem:3 Solve the following equation for $x$

$(\tan ^{-1} 2 x)$+ $(\tan ^{-1} 3 x)=\frac{\pi}{4}, x>0 .$

Solution:

$\tan ^{-1} 2 x =(\frac{\pi}{4})-\tan ^{-1} 3 x$

$=\tan ^{-1} 1-\tan ^{-1} 3 x$

$\tan ^{-1} 1+\tan ^{-1}(-3 x)$

$\tan ^{-1}(2 x)=\tan ^{-1}(\frac{1-3 x}{1+3 x})$

$\tan ^{-1} a =\tan ^{-1} b$

$a = b$

$\tan ^{-1}(-x) =-\tan ^{-1}(x)$

$\tan ^{-1} x+\tan ^{-1} y= \begin{cases}\tan ^{-1}\left(\frac{x+y}{1-x y}\right), x y<1 \\ \pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right), \begin{array}{l}x>0, y>0, x y>1\end{array} \\ -\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right), x<0, y<0, x y>1\end{cases}$

Problem:4

If $x, y$ and $z$ are in A.P. and $\tan ^{-1} x, \tan ^{-1} y$ and $\tan ^{-1} z$ are also in A.P., then

a) $x=y=z$

b) $2 x=3 y=6 z$

e) $6 x=3 y=2 z$

d) $6 x=4 y=3 z$

Solution:

$y=\frac{x+z}{2}, \quad y-x=z-y$

$\underbrace{\tan ^{-1} y-\tan ^{-1} x}_{\theta \geqslant 0}=\tan ^{-1} z-\tan ^{-1} y$

$2\theta<\pi$

$0 \leqslant \theta < \frac{\pi}{2}$

$\underbrace{\tan ^{-1} y-\tan ^{-1} x}_{0 ≤ \theta<\frac{\pi}{2} \text {⊆ Range set of} \tan^{-1}}=\tan ^{-1} z-\tan ^{-1} y$

$\tan (\theta) =\tan (\tan ^{-1} y - \tan^{-1} x) =\frac{y-x}{1+x y} \geqslant 0$

$\theta =\tan ^{-1} \frac{y-x}{1+x y}=\tan ^{-1} \frac{z-y}{1+z y}$

$\frac{y-x}{1+x y}=\frac{z-y}{1+z y}$

$(y-x)(1+z y)=(z-y)(1+x y)$

$y+z y^2-x-x y z=z+x y z-y-xy^2$

($\because$ $x,y,z$ are in A.P. $\quad \quad$ $\therefore$ $2y = x + z$ or $y-x = z - y)$

$2 x y z =y^2(x+z)=2 y^{3}$

$2 x y z =2 y^3$

$2 y (y^2-x z)=0$

$y=0 \text { or } y^2=x z$

For, $y =0$ ,

As, $x + z=2 y = 0 \Rightarrow z = -x$

$\tan ^{-1} x, \tan ^{-1} (-y) =0 \quad \tan ^{-1} = - \tan^{1} x$

Possibility 1 : $x, 0, -x \leftarrow$ in A.P.

$\tan^{-1} x , 0 , \tan^{-1} (-x)$ $\leftarrow$ in A.P.

For, $y^2 = xz$ , $x + z = 2y$

Possibility 2: $\Rightarrow x = y = z$

Problem:5

If $0 < x < 1$ then,

• $\sqrt{1+x^2}\sqrt{[\{x \cos(\cot ^{-1} x)+\sin(\cot ^{-1} x)\}^2-1]}$

is equal to ___ ?

Solution

$\cot ^{-1} x$ = $\theta, \theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$

$x = \cot \theta$

$\sqrt{1+\cot ^2 \theta}$

=$\sqrt{1+\frac{1}{\tan ^2 \theta}}=\sqrt{\frac{1+\tan ^2 \theta}{\tan ^2 \theta}}$

$=(\frac{1} {\sin \theta})$

$\sqrt{(1+x^2)\{x \cos \theta+\sin \theta\}^2-(1+x^2)}$

=$\sqrt{\{\sqrt{1+x^2}(x \cos \theta+\sin \theta)\}^2-(1+x^2)}$

$(\because \sqrt{1+x^2} =\frac{1}{sin \theta} )$

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

$=\sqrt{{(x \cot \theta+1)}^2-(1+x^2)}$

($\because \cot \theta = x$)

Problem:6

Find the number of positive solutions satisfying the equation

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

$x >0$

$2 x+1 >1$

$0 < \frac{1}{2 x+1} \cdot \frac{1}{4 x+1}<1$

Solution

$\tan ^{-1} x+\tan ^{-1} y= \begin{cases}\tan ^{-1}\left(\frac{x+y}{1-x y}\right), x y<1 \\ \pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right), \begin{array}{l}x>0, y>0, x y>1\end{array} \\ -\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right), x<0, y<0, x y>1\end{cases}$

$\frac{6 x+2}{[(2 x+1)(4 x+1)-1]}=\frac{2}{x^2}$

$3 x^3-7 x^2-6 x=0$

$x(3 x^2-7 x-6)=0$

$x(3 x+2)(x-3)=0$

$x = 0,- \frac{2}{3}, 3$