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
$\left(y^{2}+2\right)\left(y-\frac{\cos x}{1+\sin x}\right)=0 \Rightarrow y=\frac{\cos x}{1+\sin x}$
$=\frac{\frac{1-\tan ^{2} \frac{ x }{2}}{1+\tan ^{2} \frac{ x }{2}}}{1+\frac{2 \tan \frac{ x }{2}}{1+\tan ^{2} \frac{ x }{2}}}=\frac{1- t ^{2}}{1+ t ^{2}+2 t }=\frac{(1- t )(1+ t )}{(1+ t )^{2}}=\frac{1- t }{1+ t }=\frac{2}{1+ t }-1$
where $t =\tan \frac{ x }{2} \Rightarrow y '( t )=\frac{-2}{(1+ t )^{2}} \cdot$ At $x =\frac{\pi}{2}, t =1$
$\therefore y '(1)=\frac{-2}{(1+1)^{2}}=-\frac{1}{2}$