Question

For $\theta \in (0,\frac{\pi }{2}),{\text{sech}}^{-1}\left(\cos \theta \right)=$

• A. $\log \left|\tan (\frac{\pi }{6}+\frac{\theta }{2})\right|$
• B. $\log \left|\tan (\frac{\pi }{3}+\frac{\theta }{2})\right|$
• C. $\log \left|\tan (\frac{\pi }{4}+\frac{\theta }{2})\right|$
• D. $\log \left|\tan (\frac{\pi }{2}-\frac{\theta }{2})\right|$

Answer 1

Answer: (C)

$\log \left|\tan (\frac{\pi }{4}+\frac{\theta }{2})\right|$

We know that,

${\text{sech}}^{-1}\left(x\right)=\log \left(\frac{1+\sqrt{1-x^2}}{x}\right)$

Substituting $x=\cos \theta$

${\text{sech}}^{-1}\left(\cos \theta \right)=\log \left(\frac{1+\sqrt{1-{\cos }^2\theta }}{\cos \theta }\right)$

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

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

$=\log \left(\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}\right)$

$\therefore {\text{sech}}^{-1}\left(\cos \theta \right)=\log \left|\tan (\frac{\pi }{4}+\frac{\theta }{2})\right|$