Question

The general solution of $4{\sin }^{2}x+{\tan }^{2}x+{\text{cosec}}^{2}x+{\cot }^{2}x-6=0$ is
(where $n\in Z$)

• A. $(2n+1)\frac{\pi }{4}$
• B. $(4n+1)\frac{\pi }{3}$
• C. $\frac{n\pi }{4}$
• D. $(4n-1)\frac{\pi }{4}$

Answer 1

The correct option is A $(2n+1)\frac{\pi }{4}$

$4{\sin }^{2}x+{\tan }^{2}x+{\text{cosec}}^{2}x+{\cot }^{2}x-6=0\Rightarrow (2\sin x{)}^2+(\text{cosec}x{)}^2-4+(\tan x{)}^2+(\cot x{)}^2-2=0$
$\Rightarrow (2\sin x-\text{cosec}x{)}^2+(\tan x-\cot x{)}^2=0$
Which is possible when
$\Rightarrow 2\sin x-\text{cosec}x=0\text{and}\tan x=\cot x$
$\Rightarrow {\sin }^{2}x=\frac{1}{2}\text{and}{\tan }^{2}x=1$
$\Rightarrow \sin x=\pm \frac{1}{\sqrt{2}}\text{and}\tan x=\pm 1\Rightarrow x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\cdots$

$\therefore x=(2n+1)\frac{\pi }{4},n\in Z$