Q103 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics
If $y=\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)$ then $\frac{d y}{d x}=$
- A$\frac{(3-2 \sin x)^2}{13 \sin ^2 x-24 \sin x+13}$
- Correct answerB$\frac{-5 \cos x}{13 \sin ^2 x-24 \sin x+13}$
- C$\frac{5 \sin x}{13 \sin ^2 x-24 \sin x+13}$
- D$\frac{-5 \sin x}{13 \sin ^2 x-24 \sin x+13}$
Solution
Given, $y=\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)$
Now, $\frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)\right)$
$=\frac{1}{\left(\frac{2-3 \sin x}{3-2 \sin x}\right)^2+1} \times \frac{d}{d x}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)$
$=\frac{(3-2 \sin x)^2}{4+9 \sin ^2 x-12 \sin x+9+4 \sin ^2 x-12 \sin x} \times$
$\frac{(3-2 \sin x)(-3 \cos x)-(2-3 \sin x)(-2 \cos x)}{(3-2 \sin x)^2}$
$=\frac{-9 \cos x+4 \cos x}{13 \sin ^2 x-24 \sin x+13}$
$\Rightarrow \frac{d y}{d x}=\frac{-5 \cos x}{13 \sin ^2 x-24 \sin x+13}$
Hence, the correct answer is option: 'B'.