Inverse Trigonometric Functions Question 165
Question: $ {{\cos }^{-1}}( \frac{15}{17} )+2{{\tan }^{-1}}( \frac{1}{5} )= $
[EAMCET 1981]
Options:
A) $ \frac{\pi }{2} $
B) $ {{\cos }^{-1}}( \frac{171}{221} ) $
C) $ \frac{\pi }{4} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{\cos }^{-1}}( \frac{15}{17} )+2{{\tan }^{-1}}( \frac{1}{5} ) $
$ ={{\cos }^{-1}}( \frac{15}{17} )+{{\cos }^{-1}}( \frac{1-1/25}{1+1/25} ) $
$ ={{\cos }^{-1}}( \frac{15}{17} )+{{\cos }^{-1}}( \frac{12}{13} ) $
$ ={{\cos }^{-1}}( \frac{15}{17}\times \frac{12}{13}-\sqrt{1-{{( \frac{15}{17} )}^{2}}}\sqrt{1-{{( \frac{12}{13} )}^{2}}} ) $
$ ={{\cos }^{-1}}( \frac{140}{221} ) $ .
BETA
