Question

If y = tan^–1$\sqrt{\frac{1 - \sin x}{1 + \sin x}}$, then the value of dy/dx at x = π/6 is

• A. –1/2
• B. ½
• C. 1
• D. –1

Answer 1

Answer: (A)

–1/2

Answer 2

y = tan^–1 $\sqrt{\frac{(\sin x/2 - \cos x/2)^{2}}{(\sin x/2 + \cos x/2)^{2}}}$

= tan^–1 $\left| \frac{\sin x/2 - \cos x/2}{\sin x/2 + \cos x/2} \right|$

x = π/6, sinx/2 – cos x/2 < 0

∴ $\left| \frac{\sin x/2 - \cos x/2}{\sin x/2 + \cos x/2} \right|$ = – $\frac{\sin x/2 - \cos x/2}{\sin x/2 + \cos x/2}$

= $\frac{1 - \tan x/2}{1 + \tan x/2}$

= tan $\left( \frac{\pi}{4} - \frac{x}{2} \right)$

y = $\frac{\pi}{4} - \frac{x}{2}$ ⇒ $\frac{dy}{dx}$ = – $\frac{1}{2}$