Question

If $\pi <\alpha <\frac{3\pi }{2}$ then $\sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}+\sqrt{\frac{1+\cos \alpha }{1-\cos \alpha }}$ is equal to

• A. $\frac{2}{\sin \alpha }$
• B. $-\frac{2}{\sin \alpha }$
• C. $\frac{1}{\sin \alpha }$
• D. $-\frac{1}{\sin \alpha }$

Answer 1

Answer: (B)

$-\frac{2}{\sin \alpha }$

Let, $y=\sqrt{\frac{1-cos\alpha }{1+cos\alpha }}+\sqrt{\frac{1+cos\alpha }{1-cos\alpha }}$
$=\sqrt{\frac{2si{n}^2\left(\frac{\alpha }{2}\right)}{2co{s}^2\left(\frac{\alpha }{2}\right)}}+\sqrt{\frac{2co{s}^2\left(\frac{\alpha }{2}\right)}{2si{n}^2\left(\frac{\alpha }{2}\right)}}=\mid tan\left(\frac{\alpha }{2}\right)\mid +\mid \frac{1}{tan\left(\frac{\alpha }{2}\right)}\mid$

but given $\pi <\alpha <\frac{3\pi }{2}\Rightarrow \frac{\pi }{2}<\alpha <\frac{3\pi }{4}\Rightarrow tan\left(\frac{\alpha }{2}\right)<0$
Hence $y=-tan\left(\frac{\alpha }{2}\right)-\frac{1}{tan\left(\frac{\alpha }{2}\right)}=-\left(\frac{1+ta{n}^2\left(\frac{\alpha }{2}\right)}{tan\left(\frac{\alpha }{2}\right)}\right)=-\frac{2}{sin\alpha }$

Hence, option 'B' is correct.