Question

General solution of equation $\cot \theta +\text{cosec}\theta =\sqrt{3}$ is

• A. $2n\pi +\frac{\pi }{4}$
• B. $\left(2n-1\right)\pi$
• C. $2n\pi +\frac{\pi }{3}$
• D. $2n\pi +\frac{\pi }{6}$

Answer 1

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

Given expression:

$\cot \theta +\text{cosec}\theta =\sqrt{3}$

$\frac{\cos \theta }{\sin \theta }+\frac{1}{\sin \theta }=\sqrt{3}$

Taking square both side,

${\cos }^2\theta +1+2\cos \theta =3-3{\cos }^2\theta$ [\sin ce ${\sin }^2\theta =1-{\cos }^2\theta$]

$4{\cos }^2\theta +2\cos \theta -2=0$

$4{\cos }^2\theta +4\cos \theta -2\cos \theta -2=0$

We get

$(\cos \theta +1)(4\cos \theta -2)=0$

When,

$(\cos \theta +1)=0$

$\cos \theta =-1$

$\cos \pi =\cos (2n+1)\pi =-1$

Hence $\theta =(2n+1)\pi .$

When,

$(4\cos \theta -2)=0$

$\cos \theta =\frac{1}{2}$

$\cos \frac{\pi }{3}=\cos (2n\pi \pm \frac{\pi }{3})=\frac{1}{2}$

Hence, $\theta =(2n\pi \pm \frac{\pi }{3})$