Question

Let S = {$x\in (-\pi ,\pi );x\ne 0,\pm \frac{\pi }{2}$}. The sum of all distinct solution of the equation $\sqrt{3}secx+cosecx+2(tanx-cotx)=0$ in the set S is equal to-

• A. $-\frac{7\pi }{9}$
• B. $-\frac{2\pi }{9}$
• C. $0$
• D. $-\frac{5\pi }{9}$

Answer 1

The correct option is C $0$

$\sqrt{3}\sec x+\csc x+2(\tan x-\cot x)=0$

$\Rightarrow \sqrt{3}\sec x+\csc x=2(\cot x-\tan x)$

$\Rightarrow \frac{(\sqrt{3}\sec x+\csc x)}{2}=\cot x-\tan x$

$\Rightarrow \frac{\sqrt{3}}{2}\sec x+\frac{1}{2}\csc x=\cot x-\tan x$ by dividing both sides by $2$

We know that $\sec x=\frac{1}{\cos x},\csc x=\frac{1}{\sin x},\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$

$\Rightarrow \frac{\sqrt{3}}{2}\frac{1}{\cos x}+\frac{1}{2}\frac{1}{\sin x}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$

$\Rightarrow \frac{\sqrt{3}}{2}\frac{\sin x}{\sin x\cos x}+\frac{1}{2}\frac{\cos x}{\sin x\cos x}=\frac{{\cos }^{2}x}{\sin x\cos x}-\frac{{\sin }^{2}x}{\cos x\sin x}$

$\Rightarrow \frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x={\cos }^{2}x-{\sin }^{2}x$

$\Rightarrow \sin \frac{\pi }{3}\sin x+\cos \frac{\pi }{3}\cos x=\cos 2x$ where $\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}$ and $\frac{1}{2}=\cos \frac{\pi }{3}$

$\Rightarrow \cos (\frac{\pi }{3}-x)=\cos 2x$

$\Rightarrow 2x=2n\pi \pm (x-\frac{\pi }{3})$ since if $\cos \theta =\cos \alpha \Rightarrow \theta =2n\pi \pm \alpha$

$\Rightarrow 2x=2n\pi +x-\frac{\pi }{3}$ or $2x=2n\pi -x+\frac{\pi }{3}$

$\Rightarrow x=2n\pi -\frac{\pi }{3}$ or $3x=2n\pi +\frac{\pi }{3}$ or $x=\frac{2n\pi }{3}+\frac{\pi }{9}$

For $n=0,x=\frac{-\pi }{3},\frac{\pi }{9}$

For $n=-1,x=\frac{-7\pi }{3}$ which does not lie between $(-\pi ,\pi )$

and $x=\frac{-2\pi }{3}+\frac{\pi }{9}=\frac{-5\pi }{9}\in (-\pi ,\pi )$

For $n=2,x=\frac{5\pi }{3}$ does not lie between $(-\pi ,\pi )$

and $x=\frac{2\pi }{3}+\frac{\pi }{9}=\frac{7\pi }{9}\in (-\pi ,\pi )$

$\therefore x=\frac{-\pi }{3},\frac{\pi }{9},\frac{-5\pi }{9},\frac{7\pi }{9}$

Sum of all the solutions$=\frac{-\pi }{3}+\frac{\pi }{9}-\frac{5\pi }{9}+\frac{7\pi }{9}=\frac{-3\pi }{9}+\frac{\pi }{9}-\frac{5\pi }{9}+\frac{7\pi }{9}=0$