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Q. If $\sin^{-1}\left(\frac{x}{13}\right)+cosec^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2}$ , then the value of x is

Inverse Trigonometric Functions

Solution:

$sin^{-1}\left(\frac{x}{13}\right) = \frac{\pi}{2}-cosec^{-1}\left(\frac{13}{12}\right)$
$ \quad $[From given equation]
$ = sec^{-1}\left(\frac{13}{12}\right) = cos^{-1}\left(\frac{12}{13}\right) = sin^{-1} \frac{5}{13}$
$\therefore \frac{x}{13} = \frac{5}{13} $
$\Rightarrow x= 5$