Question

If $4\cos \theta -3\sec \theta =2\tan \theta$, then $\theta$ is equal to

• A. $n\pi +(-1{)}^n\frac{\pi }{10}$
• B. $n\pi +(-1{)}^n\frac{\pi }{6}$
• C. $n\pi +(-1{)}^n\frac{3\pi }{10}$
• D. $n\pi$

Answer 1

The correct option is A $n\pi +(-1{)}^n\frac{\pi }{10}$

Given,

$4\cos \theta -3\sec \theta =2\tan \theta$

$4\cos \theta -3\times \frac{1}{\cos \theta }=2\times \frac{\sin \theta }{\cos \theta }$

multiply by $\cos \theta$

$4{\cos }^2\theta -3=2\sin \theta$...........$\left(i\right)$

$\because ({\cos }^2\theta =1-{\sin }^2\theta )$

$4-4{\sin }^2\theta -3-2\sin \theta =0$............$from\left(i\right)$

$4{\sin }^2\theta +2\sin \theta -1=0$

$\therefore \sin \theta =\frac{-1\pm \sqrt{5}}{4}$

$\Rightarrow \theta =n\pi +(-1{)}^n\frac{\pi }{10}$