Question

If $\cot \theta -\tan \theta =\sec \theta$ , then $\theta$ is equal to

• A. $n\pi +(-1{)}^n\frac{\pi }{6}$
• B. $n\pi +\frac{\pi }{2}$
• C. $2n\pi +3\frac{\pi }{2}$
• D. none of these.

Answer 1

Answer: (A)

$\cot \theta -\tan \theta =\sec \theta$
$\Rightarrow$ $\frac{\cos \theta }{\sin \theta }-\frac{\sin \theta }{\cos \theta }=\frac{1}{\cos \theta }$
$[\cos \theta \ne 0,\sin \theta \ne 0]$
$\Rightarrow$ ${\cos }^2\theta -{\sin }^2\theta =\sin \theta$
$\Rightarrow$ $1-2{\sin }^2\theta =\sin \theta$
$\Rightarrow$ $2{\sin }^2\theta +\sin \theta -1=0$
$\Rightarrow$ $(2\sin \theta -1)(\sin \theta +1)=0$
$\Rightarrow$ $\sin \theta =\frac{1}{2}=\sin \frac{\pi }{6}$ $[\because \sin \theta \ne -1]$
$\Rightarrow$ $\theta =n\pi +(-1{)}^n\frac{\pi }{6}$