Question

Let $0<x<\pi$ and $y(x)$ be given by $(1+\sin x)y^3-(\cos x)y^2+2(1+\sin x)y-2\cos x=0$ The derivative of $y$ with respect to $\tan \frac{x}{2}$ at $x=\frac{\pi }{2}$ is :

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. 2
• D. -2

Answer 1

Answer: (B)

The given eq. can be written as
$\left(y^2+2\right)\left(y-\frac{\cos x}{1+\sin x}\right)=0\Rightarrow y=\frac{\cos x}{1+\sin x}$
$=\frac{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}{1+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}=\frac{1-t^2}{1+t^2+2t}=\frac{(1-t)(1+t)}{(1+t{)}^2}=\frac{1-t}{1+t}=\frac{2}{1+t}-1$
where $t=\tan \frac{x}{2}\Rightarrow y^{\prime }(t)=\frac{-2}{(1+t{)}^2}\cdot$ At $x=\frac{\pi }{2},t=1$
$\therefore y^{\prime }(1)=\frac{-2}{(1+1{)}^2}=-\frac{1}{2}$