Question

The solution set of the trigonometric equation $\tan \theta +5\cot \theta =\sec \theta$ is

• A. $\left\{\frac{\theta }{\theta }=2n\pi \pm \frac{\pi }{3},n\in Z\right\}$
• B. $\left\{\frac{\theta }{\theta }=n\pi +(-1{)}^n\frac{\pi }{2},n\in Z\right\}$
• C. $\left\{\frac{\theta }{\theta }=n\pi +\frac{\pi }{6},n\in Z\right\}$
• D. $\phi$

Answer 1

Answer: (D)

We have
$\tan \theta +5\cot \theta =\sec \theta$
$\Rightarrow \frac{\sin \theta }{\cos \theta }+\frac{5\cos \theta }{\sin \theta }=\frac{1}{\cos \theta },\sin \theta \ne 0,\cos \theta \ne 0$
$\Rightarrow \frac{{\sin }^2\theta +5{\cos }^2\theta }{\cos \theta \sin \theta }=\frac{1}{\cos \theta },\sin \theta \ne 0,\cos \theta \ne 0$
$\Rightarrow {\sin }^2\theta +5\left(1-{\sin }^2\theta \right)=\sin \theta ,\sin \theta \ne 0,\cos \theta \ne 0$
$\Rightarrow 4{\sin }^2\theta +\sin \theta -5=0,\sin \theta \ne 0,\cos \theta \ne 0$
$\Rightarrow (4\sin \theta +5)(\sin \theta -1)=0,\sin \theta \ne 0,\cos \theta \ne 0$
$\sin \theta =\frac{-5}{4},1,\sin \theta \ne 0,\cos \theta \ne 0$
$\Rightarrow \sin \theta =1,\sin \theta \ne 0,\cos \theta \ne 0[\because -1\le \sin \theta \le 1]$
$\Rightarrow \theta \in \varphi [$ if $\sin \theta =0\Rightarrow \cos \theta =0]$