Inverse Trigonometric Functions Question 12
Question: If $ A={{\tan }^{-1}}x $ , then $ \sin 2A= $
[MNR 1988; UPSEAT 2000]
Options:
A) $ \frac{2x}{\sqrt{1-x^{2}}} $
B) $ \frac{2x}{1-x^{2}} $
C) $ \frac{2x}{1+x^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Given that $ A={{\tan }^{-1}}x $ Now $ x=\tan A\Rightarrow \sin 2A=\frac{2\tan A}{1+{{\tan }^{2}}A}=\frac{2x}{1+x^{2}} $ .
BETA
