Question

Solve the equation
$tan(x+a)tan(x+b)+tan(x+b)tan(x+c)+$ $tan(x+c)tan(x+a)=1$.

• A. $n\pi ,n\in I$
• B. $\frac{n\pi }{3}+\frac{\pi }{6}-\left(\frac{a+b+c}{3}\right)$, $n\in I$
• C. $\frac{n\pi }{3}-\frac{\pi }{6},n\in I$
• D. None of these

Answer 1

Answer: (B)

Given equation is
$tan(x+b)tan(x+b)+tan(x+b)tan(x+c)+$
$tan(x+c)tan(x+a)=1$
$\Rightarrow tan\left(x+b\right)\left\{tan\left(x+a\right)+tan\left(x+c\right)\right\}=$
$1-tan\left(x+c\right)tan\left(x+a\right)$
$\Rightarrow tan\left(x+b\right)\left\{\frac{tan\left(x+a\right)+tan\left(x+c\right)}{1-tan\left(x+a\right)tan\left(x+c\right)}\right\}=1$
$\Rightarrow tan\left(x+a+x+c\right)=\frac{1}{tan\left(x+b\right)}=cot\left(x+b\right)$
$\Rightarrow tan\left(2x+a+c\right)=tan\left(\frac{\pi }{2}-\left(x+b\right)\right)$
$\Rightarrow 2x+a+c=n\pi +\frac{\pi }{2}-\left(x+b\right),n\in I$
$\Rightarrow 3x=n\pi +\frac{\pi }{2}-\left(a+b+c\right),n\in I$
$\Rightarrow x=\frac{n\pi }{3}+\frac{\pi }{6}-\left(\frac{a+b+c}{3}\right)$, $n\in I$.