Question:
General solution of $\tan 5 x=\cot 2 x$ is
(a) $\frac{n \pi}{7}+\frac{\pi}{2}, n \in Z$
(b) $x=\frac{n \pi}{7}+\frac{\pi}{3}, n \in Z$
(c) $x=\frac{n \pi}{7}+\frac{\pi}{14}, n \in Z$
(d) $x=\frac{n \pi}{7}-\frac{\pi}{14}, n \in Z$
Solution:
(c) $x=\frac{n \pi}{7}+\frac{\pi}{14}, n \in Z$
Given;
$\tan 5 x=\cot 2 x$
$\Rightarrow \tan 5 x=\tan \left(\frac{\pi}{2}-2 x\right)$
$\Rightarrow 5 x=n \pi+\frac{\pi}{2}-2 x$
$\Rightarrow 7 x=\mathrm{n} \pi+\frac{\pi}{2}$
$\Rightarrow x=\frac{\mathrm{n} \pi}{7}+\frac{\pi}{14}, \mathrm{n} \in \mathrm{Z}$