Question

If ${\tan }^{-1}x,{\tan }^{-1}y,{\tan }^{-1}z$ are in A.P. , then $\frac{2y}{1-y^2}$=

• A. $\frac{x-z}{1+xz}$
• B. $\frac{x+z}{1-xz}$
• C. $x+z$
• D. $xy$

Answer 1

The correct option is B $\frac{x+z}{1-xz}$

If ${\tan }^{-1}x,{\tan }^{-1}y,{\tan }^{-1}zareinA.P$

$\Rightarrow 2{\tan }^{-1}y={\tan }^{-1}x+{\tan }^{-1}z$

take \tan on both sides,

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

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

$\therefore \frac{2y}{1-y^2}=\frac{x+z}{1-xy}[\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},\tan {\tan }^{-1}x=x]$