Inverse Trigonometric Functions Question 16
Question: If $ {{\cot }^{-1}}\alpha +{{\cot }^{-1}}\beta ={{\cot }^{-1}}x, $ then $ x= $
[MP PET 1992]
Options:
A) $ \alpha +\beta $
B) $ \alpha -\beta $
C) $ \frac{1+\alpha \beta }{\alpha +\beta } $
D) $ \frac{\alpha \beta -1}{\alpha +\beta } $
Show Answer
Answer:
Correct Answer: D
Solution:
Given that $ {{\cot }^{-1}}\alpha +{{\cot }^{-1}}\beta ={{\cot }^{-1}}x $
$ \Rightarrow {{\cot }^{-1}}( \frac{\alpha \beta -1}{\alpha +\beta } )={{\cot }^{-1}}x\Rightarrow x=\frac{\alpha \beta -1}{\alpha +\beta } $ .
BETA
