MHT CET · Maths · Differentiation
If \(y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}\), then \(\frac{d y}{d x}\)
- A \(\frac{1}{1+x^2}\)
- B \(\frac{1}{\sqrt{1-x^2}}\)
- C -1
- D None of these
Answer & Solution
Correct Answer
(C) -1
Step-by-step Solution
Detailed explanation
\(
y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}
\)
Put \(a=r \cos \alpha, b=r \sin \alpha\)
\(\therefore y=\tan ^{-1}\left\{\frac{r(\cos x \cos \alpha-\sin x \sin \alpha)}{r(\sin \alpha \cos x+\cos \alpha \sin x)}\right\} \)
\( =\tan ^{-1}\left[\frac{\cos (x+\alpha)}{\sin (x+\alpha)}\right]=\tan ^{-1}[\cot (x+\alpha)] \)
\( =\tan ^{-1}\left\{\tan \left[\frac{\pi}{2}-(x+\alpha)\right]\right\}=\frac{\pi}{2}-(x+\alpha) \)
\( \therefore \frac{d y}{d x}=0-(1+0)=-1\)
y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}
\)
Put \(a=r \cos \alpha, b=r \sin \alpha\)
\(\therefore y=\tan ^{-1}\left\{\frac{r(\cos x \cos \alpha-\sin x \sin \alpha)}{r(\sin \alpha \cos x+\cos \alpha \sin x)}\right\} \)
\( =\tan ^{-1}\left[\frac{\cos (x+\alpha)}{\sin (x+\alpha)}\right]=\tan ^{-1}[\cot (x+\alpha)] \)
\( =\tan ^{-1}\left\{\tan \left[\frac{\pi}{2}-(x+\alpha)\right]\right\}=\frac{\pi}{2}-(x+\alpha) \)
\( \therefore \frac{d y}{d x}=0-(1+0)=-1\)
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