Inverse Trigonometric Functions Question 187
Question: $ \tan [ {{\cos }^{-1}}\frac{4}{5}+{{\tan }^{-1}}\frac{2}{3} ] $ =
[IIT 1983; EAMCET 1988; MP PET 1990; MNR 1992]
Options:
A) 6/17
B) 17/6
C) 7/16
D) 16/7
Show Answer
Answer:
Correct Answer: B
Solution:
$ \tan [ {{\cos }^{-1}}\frac{4}{5}+{{\tan }^{-1}}\frac{2}{3} ] $
$ =\tan [ {{\tan }^{-1}}\frac{\sqrt{( 1-\frac{16}{25} )}}{\frac{4}{5}}+{{\tan }^{-1}}\frac{2}{3} ] $
$ =\tan [ {{\tan }^{-1}}( \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}.\frac{2}{3}} ) ]=\tan .{{\tan }^{-1}}\frac{17}{6}=\frac{17}{6} $ .
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