Question

If $y=\log \tan \frac{x}{2}+{\sin }^{-1}\left(\cos x\right),\frac{dy}{dx}=$

• A. $co\sec x-1.$
• B. $cotx-1.$
• C. $secx-1.$
• D. $tanx-1.$

Answer 1

The correct option is A $co\sec x-1.$

$y=\log \tan \frac{x}{2}+{\sin }^{-1}\left(\cos x\right)=\log \tan \frac{x}{2}+{\sin }^{-1}\sin (\pi /2-x)=\log \tan \frac{x}{2}+\frac{\pi }{2}-x$

$\therefore \frac{dy}{dx}=\frac{1}{\tan \left(x/2\right)}.{\sec }^2\frac{x}{2}.\frac{1}{2}-1$

$=\frac{1}{2.\frac{\sin \left(x/2\right)}{\cos \left(x/2\right)}.{\cos }^2\frac{x}{2}}-1$

$=\frac{1}{2\sin \left(x/2\right)\cos \left(x/2\right)}-1=co\sec x-1.$