Inverse Trigonometric Functions Question 190
Question: $ {{\cot }^{-1}}\frac{3}{4}+{{\sin }^{-1}}\frac{5}{13}= $
Options:
A) $ {{\sin }^{-1}}\frac{63}{65} $
B) $ {{\sin }^{-1}}\frac{12}{13} $
C) $ {{\sin }^{-1}}\frac{65}{68} $
D) $ {{\sin }^{-1}}\frac{5}{12} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ {{\cot }^{-1}}\frac{3}{4}=\theta \Rightarrow \cot \theta =\frac{3}{4} $ and $ \sin \theta =\frac{1}{\sqrt{1+{{\cot }^{2}}\theta }}=\frac{1}{\sqrt{1+(9/16)}}=\frac{4}{5} $ Hence $ {{\cot }^{-1}}\frac{3}{4}+{{\sin }^{-1}}\frac{5}{13}={{\sin }^{-1}}\frac{4}{5}+{{\sin }^{-1}}\frac{5}{13} $
$ ={{\sin }^{-1}}[ \frac{4}{5}.\sqrt{1-\frac{25}{169}}+\frac{5}{13}.\sqrt{1-\frac{16}{25}} ] $
$ ={{\sin }^{-1}}[ \frac{4}{5}.\frac{12}{13}+\frac{5}{13}.\frac{3}{5} ] $
$ ={{\sin }^{-1}}[ \frac{48+15}{65} ]={{\sin }^{-1}}\frac{63}{65} $ .
BETA
