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)

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