Question

Evaluate: $\sec h^{-1}\left(\sin \theta \right)$

• A. $\log \left(\tan \frac{\theta }{2}\right)$
• B. $\log \left(\sin \frac{\theta }{2}\right)$
• C. $\log \left(\cos \frac{\theta }{2}\right)$
• D. $\log \left(\cot \frac{\theta }{2}\right)$

Answer 1

The correct option is D $\log \left(\cot \frac{\theta }{2}\right)$

Given that,

$\sec h^{-}1\left(\sin \theta \right)$

We know that

$\sec h^{-1}x=\log (\frac{1}{x}+\sqrt{\frac{1}{x^2}-1})$

then,

$\sec h^{-1}\left(\sin \theta \right)=\log (\frac{1}{\sin \theta }+\sqrt{\frac{1-{\sin }^2\theta }{{\sin }^2\theta }})$

$=\log (\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta })$

$=\log \left(\frac{1+\cos \theta }{\sin \theta }\right)\because \cos \theta =2{\cos }^2\frac{\theta }{2}-1$

$=\log \frac{(1+2{\cos }^2\frac{\theta }{2}-1)}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}$

$=\log \left(\frac{2{\cos }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)$

$=\log \left(\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}\right)$

$=\log \left(\cot \frac{\theta }{2}\right)$