Question

If $y = \sin^{-1} \left( \frac{\sin\alpha. \sin x}{1-\cos\alpha.\sin x}\right)$ , then y'(0) is :

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

Answer 1

Answer: (D)

$ [ y'\left(x\right) = \frac{dy}{dx} = \frac{1}{\sqrt{1- \frac{\sin^{2} \alpha \sin^{2} x}{\left(1-\cos\alpha\sin x\right)^{2}}}} $
$\frac{\left(1- \cos\alpha \sin x\right) \sin\alpha\cos x -\sin\alpha \sin x .\left(-\cos\alpha \cos x\right)}{\left(1-\cos\alpha \sin x\right)^{2}} $
$\therefore y'\left(0\right) = \frac{1}{\sqrt{1-0}} \frac{\left(1-0\right) \sin\alpha - 0}{\left(1-0\right)^{2}} \sin\alpha]$