Question

If $\pi <x<2\pi$, prove that $\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}=\cot (\frac{x}{b}+\frac{\pi }{4})$
.Find $b$

Answer 1

L.H.S. $=\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}=\frac{\sqrt{2{\cos }^2\frac{x}{2}}+\sqrt{2{\sin }^2\frac{x}{2}}}{\sqrt{2{\cos }^2\frac{x}{2}}-\sqrt{2{\sin }^2\frac{x}{2}}}$
$=\frac{\sqrt{2}\left|\cos \frac{x}{2}\right|+\sqrt{2}\left|\sin \frac{x}{2}\right|}{\sqrt{2}\left|\cos \frac{x}{2}\right|-\sqrt{2}\left|\sin \frac{x}{2}\right|}$
$=\frac{\left|\cos \frac{x}{2}\right|+\left|\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}\right|-\left|\sin \frac{x}{2}\right|}$
$=\frac{-\cos \frac{x}{2}+\sin \frac{x}{2}}{-\cos \frac{x}{2}+\sin \frac{x}{2}}$ $[\because \pi <x<2\pi ,\therefore \frac{\pi }{2}<\frac{x}{2}<\pi ]$
Thus, $\cos x/2$ is negative and $\sin x/2$ is positive.
Dividing numerator and denominator by $\sin x/2$, we get
$\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}=\frac{\cot \frac{x}{2}-1}{\cot \frac{x}{2}+1}$
$=\cot (\frac{x}{2}+\frac{\pi }{4})=$ R.H.S.
Ans: 2