Inverse Trigonometric Functions Question 221
Question: The value of $ \sin {{\cot }^{-1}}\tan {{\cos }^{-1}}x $ is equal to
[Bihar CEE 1974]
Options:
A) x
B) $ \frac{\pi }{2} $
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ {{\cos }^{-1}}x=\theta \Rightarrow x=\cos \theta \Rightarrow \sec \theta =\frac{1}{x} $
$ \Rightarrow \tan \theta =\sqrt{{{\sec }^{2}}\theta -1}=\sqrt{\frac{1}{x^{2}}-1}=\frac{1}{x}\sqrt{1-x^{2}} $
Now $ \sin {{\cot }^{-1}}\tan \theta =\sin {{\cot }^{-1}}( \frac{1}{x}\sqrt{1-x^{2}} ) $
Again, putting $ x=\sin \theta $
$ \sin {{\cot }^{-1}}( \frac{1}{x}\sqrt{1-x^{2}} )=\sin {{\cot }^{-1}}( \frac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta } ) $
$ =\sin {{\cot }^{-1}}(\cot \theta )=\sin \theta =x $ .
BETA
