Question

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

• A. $\theta =\frac{n\pi }{2}+\frac{(-1{)}^n}{2}{\sin }^{-1}\frac{3}{4}$
• B. $\theta =n\pi +(-1{)}^n{\sin }^{-1}\frac{3}{4}$
• C. $\theta =n\pi \pm {\sin }^{-1}\left(\frac{3}{4}\right)$
• D. None of these.

Answer 1

Answer: (A)

We have $\frac{1}{\cos \theta }-\frac{1}{\sin \theta }=\frac{4}{3}$
$\Rightarrow$ $\sin \theta -\cos \theta =\frac{4}{3}\sin \theta \cos \theta$
$\Rightarrow 3\left(\sin \theta -\cos \theta \right)=2\sin \theta$
$\Rightarrow 9\left({\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta \right)=4{\sin }^{2}2\theta$
$\Rightarrow 9\left(1-\sin 2\theta \right)=4{\sin }^2\theta$
$\Rightarrow \sin 2\theta =\frac{3}{4}$ or $\sin 2\theta =-3$
But $\sin 2\theta =-3$ is not possible.
$\therefore$ $\sin 2\theta =\frac{3}{4}$
$\therefore$ $2\theta =n\pi +{\left(-1\right)}^n{\sin }^{-1}\frac{3}{4}$
$\Rightarrow \theta =\frac{n\pi }{2}+\frac{{\left(-1\right)}^n}{2}sin\left(\frac{3}{4}\right),n\in I$