Question

If $tan\theta -\sqrt{2}sec\theta =\sqrt{3}$, then the general solution of $\theta$

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

Answer 1

The correct option is D $n\pi +(-1{)}^n\frac{\pi }{4}+\frac{\pi }{3}$

$\tan \theta -\sqrt{2}\sec \theta =\sqrt{3}$
$\Rightarrow \frac{\sin \theta }{\cos \theta }-\frac{\sqrt{2}}{\cos \theta }=\sqrt{3}$
$\Rightarrow \sin \theta -\sqrt{2}=\sqrt{3}\cos \theta \Rightarrow \sin \theta -\sqrt{3}\cos \theta =\sqrt{2}$
$\Rightarrow \frac{1}{2}\sin \theta -\frac{\sqrt{3}}{2}\cos \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \sin (\theta -\frac{\pi }{3})=\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =n\pi +{(-1)}^n\frac{\pi }{4}+\frac{\pi }{3}$