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Q.
If sin-1 $(\frac{1}{5})$ + sec-1(2) + 2tan-1$(\frac{1}{\sqrt{3}})$ + sec-1(5) +sin-1($\frac{1}{2}$) +2 tan-1($\sqrt{3}$)=k$\pi$, then k =
Inverse Trigonometric Functions
Solution:
The given question can be written as
$sin^{-1}\left(\frac{1}{5}\right)+sec^{-1}\left(5\right)+sec^{1}\left(2\right)+sin^{-1}\left(\frac{1}{2}\right)+2tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$+ 2\,tan^{-1}\left(\sqrt{3}\right)=k\pi,$
or $\left\{sin^{-1}\left(\frac{1}{5}\right)+cos^{-1}\left(\frac{1}{5}\right)\right\} + \left\{\sec^{-1}\left(2\right)+cosec^{-1}\left(2\right)\right\}$
$+\left\{2\,tan^{-1}\left(\frac{1}{\sqrt{3}}\right)+2\,cot^{-1}\left(\frac{1}{\sqrt{3}}\right)\right\}=k\pi$
or $\frac{\pi}{2}+\frac{\pi }{2}+2\left\{tan^{-1} \frac{1}{\sqrt{3}}+cot^{-1} \frac{1}{\sqrt{3}}\right\}=k\pi$
or $\pi+2\left(\frac{\pi }{2}\right)=k\pi$
or $\pi+\pi=k\pi$
or $2\pi = k\pi$
$k=2$