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} $
$\Rightarrow \theta = \frac{n\pi}{2 } + \frac{\left(-1\right)^{n}}{2} sin \left(\frac{3}{4}\right), n \in I$