Question

If x ∈ ($\pi$, 2$\pi$) and $\frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}-\sqrt{1-cosx}}$ =$cot(a+\frac{x}{2})$, then a is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$\sqrt{1+cosx}$ = $\sqrt{2co{s}^2\frac{x}{2}}$ = $\sqrt{2}$|cos $\frac{x}{2}$|

And $\sqrt{1-cosx}$ = $\sqrt{2si{n}^2\frac{x}{2}}$ = $\sqrt{2}$|sin$\frac{x}{2}$|

⇒ $\frac{\sqrt{1+cosx}+\sqrt{1-cosx}}{\sqrt{1+cosx}-\sqrt{1-cosx}}$ = $\frac{|cos\frac{x}{2}|+|sin\frac{x}{2}|}{|cos\frac{x}{2}|-|sin\frac{x}{2}|}$

= $\frac{-cos\left(\frac{x}{2}\right)+sin\left(\frac{x}{2}\right)}{-cos\left(\frac{x}{2}\right)-sin\left(\frac{x}{2}\right)}$

= $\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}$ = tan( $\frac{\pi }{4}$ - $\frac{x}{2}$)

= cot( $\frac{\pi }{2}$ - ( $\frac{x}{4}$ + $\frac{x}{2}$))

= cot ( $\frac{\pi }{4}$ + $\frac{x}{2}$)

∴ a = $\frac{\pi }{4}$