Question

If $\frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}=\sqrt{1-cosx}}=cot(a+\frac{x}{2}),xϵ(\pi ,2\pi )$ then a___

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is A

$\frac{\pi }{4}$

$\sqrt{1+cosx}=\sqrt{2co{s}^2\frac{x}{2}}=\sqrt{2}|cos\frac{x}{2}|\sqrt{1-cosx}=\sqrt{2si{n}^2\frac{x}{2}}=\sqrt{2}|sin\frac{x}{2}|\pi <x<2\pi \Rightarrow \frac{\pi }{2}<\frac{x}{2}<\pi .\therefore \sqrt{1+cosx}=-cos\frac{x}{2}\sqrt{1-cosx}=sin\frac{x}{2}$
$\therefore \frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}-\sqrt{1-cosx}}=\frac{-cos\frac{x}{2}+sin\frac{x}{2}}{-cos\frac{x}{2}-sin\frac{\pi }{2}}\Rightarrow \frac{-1tan\frac{x}{2}}{-1-tan\frac{x}{2}}=\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}=tan(\frac{\pi }{4}-\frac{x}{2})=cot(\frac{\pi }{2}-(\frac{\pi }{4}-\frac{x}{2}))=cot(\frac{\pi }{4}+\frac{x}{2})$