Question

The solution of $4{\sin }^{2}x+{\tan }^{2}x+{\csc }^{2}x+{\cot }^{2}x-6=0$ is

• A. $n\pi \pm \frac{\pi }{4}$
• B. $2n\pi \pm \frac{\pi }{4}$
• C. $n\pi +\frac{\pi }{3}$
• D. $n\pi -\frac{\pi }{6}$

Answer 1

The correct option is A $n\pi \pm \frac{\pi }{4}$

$4{\sin }^{2}x+{\tan }^{2}x+{\csc }^{2}x+{\cot }^{2}x-6=0$

$(2\sin x{)}^2+{\csc }^{2}x-2\cdot 2\sin x\cdot \csc x+(\tan x{)}^2+(\cot x{)}^2-2\cdot \tan x\cdot \cot x=0$

$(2\sin x-\csc x{)}^2+(\tan x-\cot x{)}^2=0$

If $a^2+b^2=0$ then $a=0$ and $b=0$

$\therefore 2\sin x-\csc x=0$-------1

and $\tan x-\cot x=0$------2

From Equation 1,

$2{\sin }^{2}x=1$

or ,$\sin x=\pm \frac{1}{\sqrt{2}}=\pm \sin \frac{\pi }{4}$

or, $x=n\pi +(-1{)}^n\frac{\pi }{4}$

$=n\pi \pm \frac{\pi }{4}$

From Equation 2,

$\tan x=\pm \tan \frac{\pi }{4}$

$x=n\pi \pm \frac{\pi }{4}$