Question

If ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4},$ then the value of $\tan ({\tan }^{-1}x+{\tan }^{-1}4x)$ is

• A. $\frac{15}{16}$
• B. $\frac{8}{15}$
• C. $\frac{11}{15}$
• D. $\frac{13}{15}$

Answer 1

The correct option is A $\frac{15}{16}$

Given, ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}\dots \left(1\right)$

If $2x,3x>0$ and $\left(2x\right)\left(3x\right)>1,$ then either $2x$ or $3x$ will be greater than $1.$ Then for equation $\left(1\right),$ L.H.S. $>\frac{\pi }{4}$ (reject it)

If $2x,3x>0$ and $\left(2x\right)\left(3x\right)<1,$ equation $\left(1\right)$ can be written as
${\tan }^{-1}\left(\frac{5x}{1-6x^2}\right)=\frac{\pi }{4}$
$\Rightarrow \frac{5x}{1-6x^2}=1$
$\Rightarrow 6x^2+5x-1=0$
$\Rightarrow x=\frac{1}{6}$
or $x=-1$ $($not acceptable because for equation $\left(1\right),$ L.H.S. becomes negative while R.H.S. is positive$)$

Now, $\tan ({\tan }^{-1}x+{\tan }^{-1}4x)$
$=\tan ({\tan }^{-1}\frac{1}{6}+\tan \frac{2}{3})=\frac{\frac{1}{6}+\frac{2}{3}}{1-\frac{1}{6}\cdot \frac{2}{3}}=\frac{15}{16}$