Inverse Trigonometric Functions Question 215
Question: If $ {{({{\tan }^{-1}}x)}^{2}}+{{({{\cot }^{-1}}x)}^{2}}=\frac{5{{\pi }^{2}}}{8}, $ then $ x $ equals
Options:
A) -1
B) 1
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{({{\tan }^{-1}}x)}^{2}}+{{({{\cot }^{-1}}x)}^{2}}=\frac{5{{\pi }^{2}}}{8} $
Therefore $ {{({{\tan }^{-1}}x+{{\cot }^{-1}}x)}^{2}}-2{{\tan }^{-1}}x( \frac{\pi }{2}-{{\tan }^{-1}}x )=\frac{5{{\pi }^{2}}}{8} $
Therefore $ \frac{{{\pi }^{2}}}{4}-2\times \frac{\pi }{2}{{\tan }^{-1}}x+2{{({{\tan }^{-1}}x)}^{2}}=\frac{5{{\pi }^{2}}}{8} $
Therefore $ 2{{({{\tan }^{-1}}x)}^{2}}-\pi {{\tan }^{-1}}x-\frac{3{{\pi }^{2}}}{8}=0 $
Therefore $ {{\tan }^{-1}}x=-\frac{\pi }{4},\frac{3\pi }{4} $
Therefore $ {{\tan }^{-1}}x=-\frac{\pi }{4}\Rightarrow x=-1 $ .
BETA
