Inverse Trigonometric Functions Question 120
Question: In a triangle ABC. If $ A={{\tan }^{-1}}2 $ and $ B={{\tan }^{-1}}3, $ then C is equal to
Options:
A) $ \frac{\pi }{3} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{6} $
D) $ \frac{\pi }{2} $
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Answer:
Correct Answer: B
We have $ A={{\tan }^{-1}}2\Rightarrow \tan A=2 $ and $ B={{\tan }^{-1}}3\Rightarrow \tan B=3. $ Since, A, B, C are angles of a triangle
$ \therefore A+B+C=\pi $
$ \Rightarrow C=\pi -(A+B) $ ? (1) Now, $ A+B={{\tan }^{-1}}2+{{\tan }^{-1}}3 $
$ =\pi +{{\tan }^{-1}}( \frac{2+3}{1-2.3} ) $
$ [ \because {{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\frac{x+y}{1-xy} ] $
$ =\pi +{{\tan }^{-1}}(-1)=\pi -ta{n^{-1}}(-1) $
$ =\pi -\frac{\pi }{4}=\frac{3\pi }{4} $
$ \therefore $ from (1), $ C=\pi -\frac{3\pi }{4}=\frac{\pi }{4}. $
BETA
