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