Inverse Trigonometric Functions Question 32
Question: The value of $ \sin ( 2{{\tan }^{-1}}( \frac{1}{3} ) )+\cos ({{\tan }^{-1}}2\sqrt{2})= $
[AMU 1999]
Options:
A) $ \frac{16}{15} $
B) $ \frac{14}{15} $
C) $ \frac{12}{15} $
D) $ \frac{11}{15} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \sin [ 2{{\tan }^{-1}}( \frac{1}{3} ) ]+\cos [{{\tan }^{-1}}(2\sqrt{2})] $ = $ \sin [ {{\tan }^{-1}}\frac{2/3}{1-1/9} ]+\cos [{{\tan }^{-1}}(2\sqrt{2})] $
$ =\sin [{{\tan }^{-1}}3/4]+\cos [{{\tan }^{-1}}2\sqrt{2}] $
$ =\frac{3}{5}+\frac{1}{3}=\frac{14}{15} $ .
BETA
