Q76 / 160 · TS EAMCET 2010 · Mathematics
If $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$, then $\frac{d y}{d x}$ is equal to
- A$\frac{a}{x^2+a^2}$
- B$\frac{2 a}{x^2+a^2}$
- Correct answerC$\frac{4 a}{x^2+a^2}$
- D$\frac{a^2}{x^2+a^2}$
Solution
$y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$
Put $\quad x=a \tan \theta$
$\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{x}{a}\right)$
$y=\cos ^{-1}\left(\frac{a^2-a^2 \tan ^2 \theta}{a^2+a^2 \tan ^2 \theta}\right)$ $+\sin ^{-1}\left(\frac{2 a^2 \tan \theta}{a^2+a^2 \tan ^2 \theta}\right)$
$\Rightarrow y=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$
$\Rightarrow y=\cos ^{-1}(\cos 2 \theta)+\sin ^{-1}(\sin 2 \theta)$
$\left[\begin{array}{l}\because \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta} \\ \sin ^2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta}\end{array}\right]$
$\Rightarrow \quad y=2 \theta+2 \theta$
$\Rightarrow \quad y=4 \theta$
$\Rightarrow \quad y=4 \tan ^{-1}\left(\frac{x}{a}\right)$
$\Rightarrow \frac{d y}{d x}=4 \cdot \frac{1}{1+\frac{x^2}{a^2}} \cdot \frac{1}{a}=4 \cdot \frac{a^2}{a^2+x^2} \cdot \frac{1}{a}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{4 a}{a^2+x^2}$
Hence, the correct answer is option: 'C'.