Inverse Trigonometric Functions Question 182
Question: $ {{\cos }^{-1}}\frac{4}{5}+{{\tan }^{-1}}\frac{3}{5}= $
Options:
A) $ {{\tan }^{-1}}\frac{27}{11} $
B) $ {{\sin }^{-1}}\frac{11}{27} $
C) $ {{\cos }^{-1}}\frac{11}{27} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\cos }^{-1}}\frac{4}{5}+{{\tan }^{-1}}\frac{3}{5}={{\tan }^{-1}}[ \frac{\sqrt{( 1-\frac{16}{25} )}}{\frac{4}{5}} ]+{{\tan }^{-1}}\frac{3}{5} $
[ Since $ {{\cos }^{-1}}x={{\tan }^{-1}}\frac{\sqrt{(1-x^{2})}}{x} ] $
$ ={{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{3}{5}={{\tan }^{-1}}( \frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4}.\frac{3}{5}} )={{\tan }^{-1}}( \frac{27}{11} ) $ .
BETA
