Inverse Trigonometric Functions Question 115
Question: If $ a_1,a_2,a_3,….a_{n} $ is an $ A.P. $ with common difference d; $ (d>0) $ then $ \tan [ {{\tan }^{-1}}( \frac{d}{1+a_1a_2} )+{{\tan }^{-1}}( \frac{d}{1+a_2a_3} )+…+ta{n^{-1}}( \frac{d}{1+{a_{n-1}}a_{n}} ) ] $ is equal to
Options:
A) $ \frac{(n-1)d}{a_1+a_{n}} $
B) $ \frac{(n-1)d}{1+a_1a_{n}} $
C) $ \frac{nd}{1+a_1a_{n}} $
D) $ \frac{a_{n}-a_1}{a_{n}+a_1} $
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Answer:
Correct Answer: B
We have, $ {{\tan }^{-1}}( \frac{d}{1+a_1a_2} )+{{\tan }^{-1}}( \frac{d}{1+a_2a_3} )+… $
$ +{{\tan }^{-1}}( \frac{d}{1+{a_{n-1}}a_{n}} ) $
$ ={{\tan }^{-1}}( \frac{a_2-a_1}{1+a_1a_2} )+{{\tan }^{-1}}( \frac{a_3-a_2}{1+a_2a_3} )+… $
$ +{{\tan }^{-1}}( \frac{a_{n}-{a_{n-1}}}{1+{a_{n-1}}a_{n}} ) $
$ =( {{\tan }^{-1}}a_2-{{\tan }^{-1}}a_1 )+( {{\tan }^{-1}}a_3-{{\tan }^{-1}}a_2 )+… $
$ +( {{\tan }^{-1}}a_{n}-{{\tan }^{-1}}{a_{n-1}} ) $
$ ={{\tan }^{-1}}a_{n}-{{\tan }^{-1}}a_1={{\tan }^{-1}}( \frac{a_{n}-a_1}{1+a_{n}a_1} ) $
$ ={{\tan }^{-1}}( \frac{(n-1)d}{1+a_1a_{n}} ) $
$ \therefore \tan [ {{\tan }^{-1}}( \frac{d}{1+a_1a_2} )+{{\tan }^{-1}}( \frac{d}{1+a_2a_3} )+… . $
$ . …+{{\tan }^{-1}}( \frac{d}{1+{a_{n-1}}a_{n}} ) ]=\frac{(n-1d)}{1+a_1a_{n}} $
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