Question

The solution of the equation $\sec \theta -\text{cosec}\theta =\frac{4}{3}$ is

• A. $\frac{1}{2}[n\pi +(-1{)}^n{\sin }^{-1}\frac{3}{4}]$
• B. $\frac{n\pi }{2}+(-1{)}^n{\sin }^{-1}\frac{3}{4}$
• C. $n\pi +(-1{)}^n{\sin }^{-1}\frac{3}{4}$
• D. None of the above

Answer 1

The correct option is A $\frac{1}{2}[n\pi +(-1{)}^n{\sin }^{-1}\frac{3}{4}]$

$\sec \theta -\text{cosec}\theta =\frac{4}{3}$
$\Rightarrow 3(\sin \theta -\cos \theta )=4\sin \theta \cos \theta$
$\Rightarrow 9(1-\sin 2\theta )=4{\sin }^{2}2\theta$
$\Rightarrow 4{\sin }^{2}2\theta +9\sin 2\theta -9=0$
$\Rightarrow \sin 2\theta =\frac{-9\pm \sqrt{81+144}}{8}=\frac{3}{4},-3$
But $\sin 2\theta \ne -3$
$\therefore \sin 2\theta =\frac{3}{4}$
$\Rightarrow \theta =\frac{1}{2}[n\pi +(-1{)}^n{\sin }^{-1}\frac{3}{4}]$.