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

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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}

 

Posted by

Divya Prakash Singh

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