Inverse Trigonometric Functions Question 95
Question: $ {{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{3}{5}-{{\tan }^{-1}}\frac{8}{19}= $
[AMU 1976, 77]
Options:
A) $ \frac{\pi }{4} $
B) $ \frac{\pi }{3} $
C) $ \frac{\pi }{6} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{3}{5}-{{\tan }^{-1}}\frac{8}{19} $
$ ={{\tan }^{-1}}[ \frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4}\times \frac{3}{5}} ]-{{\tan }^{-1}}\frac{8}{19}={{\tan }^{-1}}\frac{27}{11}-{{\tan }^{-1}}\frac{8}{19} $
$ ={{\tan }^{-1}}[ \frac{\frac{27}{11}-\frac{8}{19}}{1+\frac{27}{11}\times \frac{8}{19}} ]={{\tan }^{-1}}( \frac{425}{425} )={{\tan }^{-1}}(1)=\frac{\pi }{4} $ .
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