# 14. If $\sin\left(\sin^{-1}\frac{1}{5} + \cos ^{-1}x \right ) =1$, then find the value of $x$.

As we know the identity;

$sin^{-1} x + cos^{-1} x = \frac {\pi}{2},\ x\ \epsilon\ [-1,1]$. it will just hit you by practice to apply this.

So, $\sin\left(\sin^{-1}\frac{1}{5} + \cos ^{-1}x \right ) =1$     or    $\sin^{-1}\frac{1}{5} + \cos ^{-1}x =\sin^{-1}(1)$,

we can then write $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$,

putting in above equation we get;

$\sin^{-1}\frac{1}{5} + \frac{\pi}{2} - \sin^{-1}x =\frac{\pi}{2}$                          $\because \left [ \sin^{-1}(1)=\frac{\pi}{2} \right ]$

$\sin^{-1}x = \sin^{-1} \frac{1}{5}$

Ans.$x = \frac{1}{5}$

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