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  4. If sin-1((2α/1+α 2))+sin-1((2β /1+β 2)) =2tan-1x, then x =

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Q. If $sin^{-1}\left(\frac{2\alpha}{1+\alpha ^{2}}\right)+sin^{-1}\left(\frac{2\beta }{1+\beta ^{2}}\right) =2tan^{-1}x$, then $x =$

Inverse Trigonometric Functions

Solution:

$sin^{-1}\left(\frac{2\alpha}{1+\alpha ^{2}}\right)+sin^{-1}\left(\frac{2\beta }{1+\beta ^{2}}\right) =2tan^{-1}x$
$\Rightarrow 2tan^{-1}\alpha + 2tan^{-1} \beta = 2tan^{-1}x$
$\Rightarrow tan^{-1}\alpha + tan^{-1}\beta = tan^{-1}x$
$\Rightarrow tan^{-1} \left(\frac{\alpha + \beta}{1- \alpha\beta}\right) = tan^{-1} x$
$\Rightarrow x = \frac{\alpha + \beta }{1- \alpha \beta }$