Question

Simplify ${\tan }^{-1}\left(\frac{6x}{1-8x^2}\right)$

• A. ${\tan }^{-1}2x+{\tan }^{-1}4x$
• B. ${\tan }^{-1}2x-{\tan }^{-1}4x$
• C. $-{\tan }^{-1}2x-{\tan }^{-1}4x$
• D. $2{\tan }^{-1}2x-{\tan }^{-1}4x$

Answer 1

The correct option is A ${\tan }^{-1}2x+{\tan }^{-1}4x$

Solve

${\tan }^{-1}\left(\frac{6x}{1-8x^2}\right)$

$={\tan }^{-1}\left(\frac{2x+4x}{1-\left(2x\right)\left(4x\right)}\right)$

We know that,

${\tan }^{-1}\left(\frac{A+B}{1-AB}\right)={\tan }^{-1}A+{\tan }^{-1}B$

So,

${\tan }^{-1}\left(\frac{2x+4x}{1-\left(2x\right)\left(4x\right)}\right)={\tan }^{-1}2x+{\tan }^{-1}4x$

$={\tan }^{-1}2x+{\tan }^{-1}4x$

Hence, this is the answer.