Question

Solve
$y={\tan }^{-1}\left(\frac{\cos x}{1-\sin x}\right)$.

Answer 1

Given,

to solve ${\tan }^{-1}\left(\frac{\cos x}{1-\sin x}\right)$

Let us try to solve$\frac{\cos x}{1-\sin x}$

since as we know that

$\cos x$ $1-\sin x$

$\cos 2x={\cos }^{2}x-{\sin }^{2}x$ $\sin 2x=2\sin x\cos x$

Let $x=\frac{x}{2}$ Let $x=\frac{x}{2}$

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

$\cos x={\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}$…...$\left(1\right)$ $\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}$….$\left(2\right)$

$\therefore {\tan }^{-1}\left(\frac{\cos x}{1-\sin x}\right)={\tan }^{-1}\left[\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{1-2\sin \frac{x}{2}\cos \frac{x}{2}}\right]$…….$\left(3\right)$

Since we know that ${\sin }^{2}x+{\cos }^{2}x=1$

then, ${\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}=1$

$={\tan }^{-1}\left[\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}\right]$

$\because$ as we know that $a^2-b^2=(a+b)(a-b)$

and $(a-b{)}^2=a^2+b^2-2ab$, here $a=\cos \frac{x}{2}$, $b=\sin \frac{x}{2}$

So, now we get,

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

$\Rightarrow {\tan }^{-1}\left[\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]$

dividing by $\cos \frac{x}{2}$ we get,

we are dividing with $\cos \frac{x}{2}$ because, we need $\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}$ in terms of $\tan$,

and as we know that $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

So, that is why let us divide whole equation with $\cos \frac{x}{2}$

$\Rightarrow {\tan }^{-1}\left[\frac{\frac{\cos \frac{x}{2}}{\cos \frac{x}{2}}+\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}}{\frac{\cos \frac{x}{2}}{\cos \frac{x}{2}}-\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}}\right]$

$\Rightarrow {\tan }^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right]$

$[\because \frac{\sin \theta }{\cos \theta }=\tan \theta ]$

$\Rightarrow {\tan }^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-1.\tan \frac{x}{2}}\right]$ $[\because \tan \frac{\pi }{4}=1]$

$\Rightarrow {\tan }^{-1}\left[\frac{1-\tan \frac{x}{2}}{1-\tan \frac{\pi }{4}\tan \frac{x}{2}}\right]$

($\because$ IT is in the form of $\tan (x+y)$ as $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$)

$\Rightarrow ta{n}^{-1}\left[\frac{\tan \frac{\pi }{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi }{4}\cdot \tan \frac{x}{2}}\right]$

Here, $x=\frac{\pi }{4},y=\frac{x}{2}$, we get,

$\Rightarrow {\tan }^{-1}\left[\tan [\frac{\pi }{4}+\frac{x}{2}]\right]$

$[\because {\tan }^{-1}\frac{1}{\tan }]$

$\Rightarrow \frac{1}{\tan }\left[\tan [\frac{\pi }{4}+\frac{x}{2}]\right]=\frac{\pi }{4}+\frac{x}{2}$

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