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If $\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$,
then what is the value of $x ?$
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then what is the value of $x ?$
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1896 Upvotes
Verified Answer
The correct answer is:
$\frac{a-b}{1+a b}$
Given,
$\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
$\therefore \quad 2 \tan ^{-1} a-2 \tan ^{-1} b=2 \tan ^{-1} x$
$\Rightarrow \tan ^{-1} a-\tan ^{-1} b=\tan ^{-1} x$
$\Rightarrow \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x$
$\Rightarrow x=\frac{a-b}{1+a b}$
$\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)-\cos ^{-1}\left(\frac{1-b^{2}}{1+b^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
$\therefore \quad 2 \tan ^{-1} a-2 \tan ^{-1} b=2 \tan ^{-1} x$
$\Rightarrow \tan ^{-1} a-\tan ^{-1} b=\tan ^{-1} x$
$\Rightarrow \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x$
$\Rightarrow x=\frac{a-b}{1+a b}$
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