Question

If $3\sec \theta +2\sec \frac{\pi }{4}=2\cos \theta$, where $\theta \in (0,2\pi )$, then which of the following can be correct?

• A. $\cos 2\theta =0$
• B. $\sin 2\theta =1$
• C. $\tan \theta =1$
• D. $\tan \theta =-1$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\cos 2\theta =0$
B $\sin 2\theta =1$
C $\tan \theta =1$
D $\tan \theta =-1$

$3\sec \theta +2\sec \frac{\pi }{4}=2\cos \theta$
$\Rightarrow \frac{3}{\cos \theta }+2\sqrt{2}=2\cos \theta$
$\Rightarrow \frac{3+2\sqrt{2}\cos \theta }{\cos \theta }=2\cos \theta$
$\Rightarrow 3+2\sqrt{2}\cos \theta =2co{s}^2\theta$
$\Rightarrow 2{\cos }^2\theta -2\sqrt{2}\cos \theta -3=0$
$\cos \theta =\frac{2\sqrt{2}\pm \sqrt{{\left(2\sqrt{2}\right)}^2-4\left(3\right)\left(2\right)}}{4}$
$=\frac{2\sqrt{2}\pm 4\sqrt{2}}{4}=\frac{\sqrt{2}\pm 2\sqrt{2}}{2}$
$=\frac{3\sqrt{2}}{2},\frac{-1}{\sqrt{2}}$
$\sin ce1\le \cos \theta \le -1$
$Hence\cos \theta =\frac{-1}{\sqrt{2}}$
$\theta =\frac{\pi }{4},\frac{7\pi }{4}\because \theta \in (0,2\pi )$
$a)\cos 2\theta =1$
$\cos \left(2\frac{\pi }{4}\right)$
$=\cos \left(\frac{\pi }{2}\right)$
$=0$
$b)sin2\theta =1$
$\sin \left(2\frac{\pi }{4}\right)$
$=\sin \left(\frac{\pi }{2}\right)$
$=1$
$c)\tan \theta =1$
$\tan \left(\frac{\pi }{4}\right)$
$=1$
$d)\tan \theta =-1$
$\tan \left(\frac{7\pi }{4}\right)=\tan (2\pi -\frac{\pi }{4})$
$=-\tan \left(\frac{\pi }{4}\right)$
$=-1$