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Question:

If $\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos ^{-1} \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$, then prove that $x=\frac{a-b}{1+a b}$

 

Solution:

Let :

$a=\tan m$

$b=\tan n$

$x=\tan y$

Now,

$\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos ^{-1} \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}$

$\Rightarrow \sin ^{-1} \frac{2 \tan m}{1+\tan ^{2} m}-\cos ^{-1} \frac{1-\tan ^{2} n}{1+\tan ^{2} n}=\tan ^{-1} \frac{2 \tan y}{1-\tan ^{2} y}$

$\Rightarrow \sin ^{-1}(\sin 2 m)-\cos ^{-1}(\cos 2 n)=\tan ^{-1}(\tan 2 y) \quad\left[\because \sin 2 x=\frac{2 \tan x}{1+\tan ^{2} x}\right.$ and $\left.\cos 2 x=\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right]$

$\Rightarrow 2 m-2 n=2 y$

$\Rightarrow m-n=y$

$\Rightarrow \tan ^{-1} a-\tan ^{-1} b=\tan ^{-1} x \quad[\because a=\tan m, b=\tan n$ and $x=\tan y]$

$\Rightarrow \tan ^{-1} \frac{a-b}{1+a b}=\tan ^{-1} x \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}\right]$

$\Rightarrow \frac{a-b}{1+a b}=x$

$\therefore \frac{a-b}{1+a b}=x$

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