Question

If $\frac{\pi }{2}<\alpha <\pi$, then $\sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}+\sqrt{\frac{1+\cos \alpha }{1-\cos \alpha }}=$

• A. $2\sec \alpha$
• B. $-2\sec \alpha$
• C. $2cosec\alpha$
• D. $-2cosec\alpha$

Answer 1

Answer: (C)

$2cosec\alpha$

$\sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}+\sqrt{\frac{1+\cos \alpha }{1-\cos \alpha }}$

$=\sqrt{\frac{(1-\cos \alpha {)}^2}{{\sin }^2\alpha }}+\sqrt{\frac{(1+\cos \alpha {)}^2}{{\sin }^2\alpha }}$

$=|cosec\alpha -cot\alpha |+|cosec\alpha +\cot \alpha |$

$=2cosec\alpha$ because $\sin \alpha >0,\alpha \in (\frac{\pi }{2},\pi )$