Question

If $co{t}^{-1}x+co{t}^{-1}y+co{t}^{-1}z=\frac{\pi }{4}$, then $xy+yz+zx+x+y+z=$

• A. $1-xyz$
• B. $-1-xyz$
• C. $1+xyz$
• D. $-1+xyz$

Answer 1

The correct option is C $1+xyz$

If ${\cot }^{-1}x+{\cot }^{-1}y+{\cot }^{-1}z=\frac{\pi }{4}$
${\cot }^{-1}x+{\cot }^{-1}y=\frac{\pi }{4}-{\cot }^{-1}z$
Take $\cot$ on both sides.
$\cot ({\cot }^{-1}x+{\cot }^{-1}y)=\cot (\frac{\pi }{4}-{\cot }^{-1}z).....[\because \cot (A+B)=\frac{\cot A\cot B-1}{\cot B+\cot A}]$
$\frac{xy-1}{x+y}=\frac{z+1}{(z-1)}$
$-xy+xyz-z+1=zx+yz+x+y$
$\Rightarrow xz+yz+xy+x+y+z=1+xyz$