Inverse Trigonometric Functions Question 48
Question: The value of $ \sin {{\cot }^{-1}}\tan {{\cos }^{-1}}x, $ is
Options:
A) $ x $
B) $ \frac{1}{x} $
C) $ 1 $
D) $ 0 $
Show Answer
Answer:
Correct Answer: A
Let $ {{\cos }^{-1}}x=\theta \Rightarrow x=\cos \theta $
Or $ \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 =sinco{t^{-1}}( \frac{1}{| x |}\sqrt{1-x^{2}} ). $
Again, putting $ x=\sin \theta , $ we get $ \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
