We are looking for something like,

$\cos ^{-1} x=\tan ^{-1}(?), \quad|x| \leqslant 1$

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

If $x \geqslant 0, \theta \in [0, \frac{\pi}{2}] \subset [-\frac{\pi}{2}, \frac{\pi}{2}].$

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

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

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

$\because x= \cos \theta$ ; $\sin^2\theta + \cos^2\theta=1$

$\Rightarrow \tan (\theta)=\frac{\sqrt{1-x^2}}{x}$ $[\because \theta \in$ Range set of $\tan ^{-1}]$

$\theta=\cos ^{-1} x=\tan ^{-1} \frac{\sqrt{1-x^2}}{x}, x \geqslant 0$

If $-1 \leqslant x<0, \theta=\cos ^{-1} x \in\left(\frac{\pi}{2}, \pi\right]$

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

$\theta ∉$ Range set of $\tan ^{-1}$

$\therefore \theta \neq \tan ^{-1} \frac{\sqrt{1-x^2}}{x}$

But, $\tan (\theta)=\tan (\theta-\pi),$

$\theta-\pi \in (-\frac{\pi}{2}, 0]$

$\therefore \theta-\pi \in$ Range set of $\tan ^{-1}$.

$\tan (\theta)=\tan (\theta-\pi)=\frac{\sqrt{1-x^2}}{x}$

$\theta-\pi \in$ Range set of $\tan ^{-1}$

$\therefore \quad \theta-\pi =\tan ^{-1} \frac{\sqrt{1-x^2}}{x}$

$\theta=\pi+\tan ^{-1} \frac{\sqrt{1-x^2}}{x}$ $\quad x<0$

So,

$\cos^{-1} x = \tan^{-1} (\frac{\sqrt{1 - x^2}}{x}) ,~ ~ x \geq 0$

$\cos^{-1} x = \pi + \tan^{-1} (\frac{\sqrt{1 - x^2}}{x}) ,~ ~ x < 0$

Now, we have to make an equation like

$\tan ^{-1} x=\cos ^{-1}(?), x \in \mathbb{R}$

Let, $x=\tan \theta \Rightarrow \tan ^{-1} x=\theta, \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

$\cos \left(\tan ^{-1} x\right)=\cos \theta=\frac{1}{\sec \theta}=\frac{1}{\sqrt{sec^{2}\theta}}$

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

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

$\theta=\cos ^{-1} \frac{1}{\sqrt{1+x^2}},$ true only when $\theta \in[0, \pi]$

We further divide into cases to get respective formulas,

case I: $x \geqslant 0$

$\theta=\tan ^{-1} x, \quad \theta \in [0, \frac{\pi}{2})$

$[0, \frac{\pi}{2}) \subset [0, \pi]$

$\leftarrow$ Range set of $\cos ^{-1}$

$\theta \in \text { Range set } {\cos ^{-1}}$

$\cos (\theta)=1 / \sqrt{1+x^2},$

$\theta=\tan ^{-1} x=\cos ^{-1} \frac{1}{\sqrt{1+x^2}}, x \geqslant 0$

case II: $x<0$

$\theta=\tan ^{-1} x, \theta \in {(\frac{-\pi}{2},{0})}$

$\theta \notin[0, \pi], \theta \notin \text { Range set of }\cos^{-1}$

$\theta+\pi \in\left(\frac{\pi}{2}, \pi\right) \subset[0, \pi]$

$\therefore \quad \theta+\pi \in \text { Range set } of \cos ^{-1}$

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

$\theta+\pi=\cos ^{-1} \frac{-1}{\sqrt{1+x^2}}$

$\theta =-\pi+\cos ^{-1}\frac{-1}{\sqrt{1+x^2}}$

$\rightarrow \cot ^{-1} x= \tan ^{-1}(1 / x), x> 0$

$\rightarrow \cot ^{-1} x=\pi+\tan ^{-1}\left(\frac{1}{x}\right), x<0$

$\rightarrow \tan ^{-1} x= \cot ^{-1}\left(\frac{1}{x}\right), x > 0$

$\rightarrow \tan ^{-1} x= \cot ^{-1}\left(\frac{1}{x}\right)-\pi, x<0$

Find

$\cot \{\sum_{n=1}^{23} \cot ^{-1} (1+\sum_{k=1}^n 2 k)\}$

Solution:

$\cot ^{-1} \left(1+\sum_{k=1}^n 2 k\right)=\cot ^{-1}(1+2 \times(1+2+.......+n))$

$=\cot ^{-1}\left(1+2 \times \frac{n(n+1)}{2}\right)$

$=\cot ^{-1}(1+n(n+1))$

= $\tan^{-1}(n+1)-\tan^{-1}(n)$

Let, $\cot ^{-1}(1+n(n+1))=\theta \in(0, \pi)$

$\cot \theta=1+n(n+1)$

$\tan \theta=\frac{1}{1+n(n+1)}=\frac{(n+1)-n}{1+(\operatorname{n+1}) n}$

$\because \tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}$

$\Rightarrow \tan \theta =\frac{\tan (\tan ^{-1} (n+1))-\tan (\tan ^{-1} n)}{1+\tan (\tan ^{-1}(n+1)) \tan (\tan ^{-1}n)}$

$\Rightarrow \tan \theta =\tan (\tan ^{-1}(n+1)-\tan ^{-1} n)$

($\because n \in N ; n, n+1 > 0$)

$\tan ^{-1}(n+1) \in(0, \pi / 2)$

$\tan ^{-1}(n) \in(0, \pi / 2)$

$\tan ^{-1}(n+1)-\tan ^{-1} n \in(-\pi / 2, \pi / 2)$ $\leftarrow$ Range set of $\tan^{-1}$

$\theta \in(0, \pi / 2)$

$[\because \theta \in (0,\pi)~$ & $~ (\frac{-\pi}{2} , \frac{\pi}{2}) ]$

$\tan ^{-1}(n+1)-\tan ^{-1} n \in (0, \pi / 2)$

$[\because \tan^{-1}$ is increasing function in $(0,\frac{\pi}{2})]$

$\Rightarrow \theta=\tan ^{-1}(n+1)-\tan ^{-1}(n)$

$[n+1>n \Rightarrow \tan^{-1}(n+1)> \tan^{-1}n]$

$\Rightarrow \cot ^{-1} \left(1+\sum_ {k=1}^n 2 k\right)= \sum_{n=1}^{23}\left(\tan ^{-1} (n+1)-\tan ^{-1} n\right)$

$= (\tan ^{-1} 2-\tan ^{-1} 1)+(\tan ^{-1}3 -\tan ^{-1} 2) \cdots +\left(\tan ^{-1}23-\tan ^{-1}22) +\left(\tan ^{-1} 24-\operatorname{tan}^{-1}23\right)\right.$

$= \tan ^{-1} 24-\tan ^{-1} 1$

$\tan ^{-1} 24-\tan ^{-1} 1$

$=\tan ^{-1} \underbrace{24}_x+\tan ^{-1} \underbrace{(-1)}_y$

$\tan ^{-1} \frac{24+(-1)}{1-(24)(-1)}=\tan ^{-1} \frac{23}{25}$

Now, according to the question,

$\cot \left( \sum_{n=1}^{23} \cot ^{-1}\left(1+\sum_{k=1}^{2 n} 2 k\right)\right)=\cot \left( \tan^{-1}\frac{23}{25} \right)$

($\because$ $\tan ^{-1} \frac{23}{25}=\theta$ ,$\Rightarrow$ $\frac{23}{25}=\tan \theta$ , $\Rightarrow$ $\cot \theta=25 / 23$)

$=\cot (\theta)$

$=25 / 23$

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

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

Problem

Find $x, 0<|x|<\sqrt{2}$

such that

Solution:

Observe,

$\ x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+\cdot \cdot \cdot$

$=x\left(1-\frac{x}{2}+\frac{x^2}{4}-\frac{x^3}{8}+\cdots\right)$

$=x\left(1+\left(-\frac{x}{2}\right)+\left(\frac{-x}{2}\right)^2+\left(-\frac{x}{2}\right)^3\cdots\right.)$

$=\frac{x}{1-\left(\frac{-x}{2}\right)} \quad$

$=\frac{x}{1+\frac{x}{2}} \quad$

$\left[ \because |x| < \sqrt{2} \Rightarrow |x| < 2 \Rightarrow |\frac{x}{2}|< 1 \Rightarrow |\frac{-x}{2}| < 1 \right]$

And,

$\ x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots$

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

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

$=\frac{x^2}{1+\frac{x^2}{2}} \left[\because |x| < \sqrt{2}, x^2 < 2, \frac{x^2}{2}<1, \left| -\frac{x^2}{2} \right| <1 \right]$

We are given,

$\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots\right)+\cos ^{-1}({x^2}-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\frac{\pi}{2}$

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

$\quad \quad [x-\frac{x^2}{2}+\frac{x^3}{4}-\cdots] = {\frac{x}{1+\frac{x}{2}}}]$

$\Rightarrow \frac{x}{1+\frac{x}{2}} =\frac{x^2}{1+\frac{x^2}{2}},$

$\Rightarrow x\left(1+\frac{x^2}{2}\right) =x^2\left(1+\frac{x}{2}\right)$

$\Rightarrow x+\frac{x^2}{2} \ =x^2+\frac{x^2}{2}$

$x (x-1)=0 \quad \because 0< |x| <\sqrt{2}$

$\Rightarrow x=1$