Inverse Trigonometric Functions Question 53
Question: If $ {{\sin }^{-1}}x={{\tan }^{-1}}y, $ what is the value of $ \frac{1}{x^{2}}-\frac{1}{y^{2}}? $
Options:
A) 1
B) -1
C) 0
D) 2
Show Answer
Answer:
Correct Answer: A
Let, $ {{\sin }^{-1}}x={{\tan }^{-1}}y=\theta $
$ \Rightarrow x=\sin \theta $ and $ y=\tan \theta $
$ \frac{1}{x^{2}}=\frac{1}{{{\sin }^{2}}\theta }=\cos ec^{2}\theta $ And $ \frac{1}{y^{2}}=\frac{1}{{{\tan }^{2}}\theta }={{\cot }^{2}}\theta . $
$ \Rightarrow \frac{1}{x^{2}}-\frac{1}{y^{2}}=\cos ec^{2}\theta -{{\cot }^{2}}\theta =1 $
BETA
