Inverse Trigonometric Functions Question 26
Question: $ 3{{\tan }^{-1}}a $ is equal to
[MP PET 1993]
Options:
A) $ {{\tan }^{-1}}\frac{3a+a^{3}}{1+3a^{2}} $
B) $ {{\tan }^{-1}}\frac{3a-a^{3}}{1+3a^{2}} $
C) $ {{\tan }^{-1}}\frac{3a+a^{3}}{1-3a^{2}} $
D) $ {{\tan }^{-1}}\frac{3a-a^{3}}{1-3a^{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
It is obvious.
$ 3{{\tan }^{-1}}a $ is equal to $ {{\tan }^{-1}}\frac{3a-a^{3}}{1-3a^{2}} $
BETA
