Inverse Trigonometric Functions Question 142
Question: $ \sin ({{\cot }^{-1}}x) $
[MNR 1987; MP PET 2001; DCE 2002]
Options:
A) $ \sqrt{1+x^{2}} $
B) $ x $
C) $ {{(1+x^{2})}^{-3/2}} $
D) $ {{(1+x^{2})}^{-1/2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ {{\cot }^{-1}}x=\theta \Rightarrow x=\cot \theta $ Now $ \cos ec\theta =\sqrt{1+{{\cot }^{2}}\theta }=\sqrt{1+x^{2}} $
$ \therefore \sin \theta =\frac{1}{\cos ec\theta }=\frac{1}{\sqrt{1+x^{2}}}\Rightarrow \theta ={{\sin }^{-1}}\frac{1}{\sqrt{1+x^{2}}} $
Hence $ \sin ({{\cot }^{-1}}x)=\sin ( {{\sin }^{-1}}\frac{1}{\sqrt{1+x^{2}}} ) $
$ =\frac{1}{\sqrt{1+x^{2}}}={{(1+x^{2})}^{-1/2}} $ .
BETA
