Question

If $tan^{-1} x + tan^{-1} y =\frac{\pi}{4} \ xy\ \textless \ 1$,then write the value of x + y + xy.

Answer 1

$\underline{\text{Answer :}}$

$\boxed{\mathrm{x+y+xy=1}}$

$\underline{\text{Step by step solution :}}$

$\mathrm{Now,\:tan^{-1}x+tan^{-1}y=\frac{\pi}{4}}$

$\to \mathrm{tan^{-1}\frac{x+y}{1-xy}=\frac{\pi}{4}}$

$\to \mathrm{\frac{x+y}{1-xy}=tan\frac{\pi}{4}}$

$\to \mathrm{\frac{x+y}{1-xy}=1}$

$\to \mathrm{x+y=1-xy}$

$\implies \boxed{\mathrm{x+y+xy=1}}$

$\underline{\text{Rules :}}$

$\mathrm{1.\:tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}}$

$\mathrm{2.\:tan^{-1}(tan\theta)=\theta}$

$\mathrm{3.\:tan(tan^{-1}\theta)=\theta}$

$\mathrm{4.\:tan\frac{\pi}{4}=1}$

Answer 2

$\huge{\mathfrak{Solution:-}}$

Given:-

$\bold{tan^{-1}x+tan^{-1}y = \frac{\pi}{4}xy\;\;\ < 1}$

Find:-

x + y + xy

So,

$\bold{tan^{-1}+tan^{-1}y = \frac{\pi}{4}}$

$\bold{\implies tan^{-1} (\frac{x+y}{1-xy}) = \frac{\pi}{4}}$

$\bold{\implies tan^{-1}[tan^{-1}(\frac{x+y}{1-xy})]=tan(\frac{\pi}{4})}$

$\bold{\implies \frac{x+y}{1-xy}= 1}$

$\bold{\implies x+y = 1-xy}$

$\bold{\implies x+y+xy = 1}$

$\boxed{\bold{x+y+xy = 1}}$