Question

Find the value of $\theta$ ,

$\frac{\cos \theta }{1-\sin \theta }+\frac{\cos \theta }{1+\sin \theta }=4$ such that $\theta \in [0,2\pi ]$

Answer 1

$\frac{\cos \theta }{1-\sin \theta }+\frac{\cos \theta }{1+\sin \theta }=4$

$\Rightarrow \cos \theta (\frac{1}{1-\sin \theta }+\frac{1}{1+\sin \theta })=4$

$\Rightarrow \cos \theta \left(\frac{1+\sin \theta +1-\sin \theta }{(1-\sin \theta )(1-\sin \theta )}\right)=4$

$\Rightarrow \cos \theta \left(\frac{2}{1-{\sin }^2\theta }\right)=4$

$\Rightarrow \cos \theta \left(\frac{2}{{\cos }^2\theta }\right)=4$

$\Rightarrow \frac{2\cos \theta }{{\cos }^2\theta }=4$

$\Rightarrow \frac{2}{\cos \theta }=4$

$\Rightarrow 2\sec \theta =4$(given)

$\Rightarrow \sec \theta =2$

$\Rightarrow \cos \theta =\frac{1}{2}$

$\cos \theta$ is positive in first and fourth quadrant.

$\therefore \cos \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3},2\pi -\frac{\pi }{3}$

$\therefore \theta =\frac{\pi }{3},\frac{5\pi }{3}$