Differentiation Question 341
Question: If $ y={{\tan }^{-1}}( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} ) $ then $ \frac{dy}{dx}= $
[Kerala (Engg.) 2005]
Options:
A) 2
B) - 1
C) $ \frac{a}{b} $
D) 0
E) $ \frac{b}{a} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ y={{\tan }^{-1}}( \frac{a\cos x-b\sin x}{b\cos x+a\sin x} ) $
Let $ a=r\sin \theta $ and $ b=r\cos \theta $
\ $ y={{\tan }^{-1}}[ \frac{r\sin (\theta -x)}{r\cos (\theta -x)} ] $
$ y=\theta -x $ ; $ y={{\tan }^{-1}}( \frac{a}{b} )-x $
$ \frac{dy}{dx}=-1 $ .
BETA
