Question

The general values of '$\theta$' satisfying the equation $\sec 4\theta -\sec 2\theta =2$ is

Answer 1

$\sec 4\theta -\sec 2\theta =2$

$\frac{1}{\cos 4\theta }-\frac{1}{\cos 2\theta }=2$

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

$\Rightarrow \cos 2\theta -\cos 4\theta =2\cos 2\theta \cos 4\theta$

$\Rightarrow \cos 2\theta -\cos 4\theta =\cos 6\theta +\cos 2\theta$ using $2\cos A\cos B=\cos (A+B)+\cos (A-B)$

$\Rightarrow -\cos 4\theta =\cos 6\theta$

$\Rightarrow -\cos 4\theta -\cos 6\theta =0$

$\Rightarrow \cos 4\theta +\cos 6\theta =0$

$\Rightarrow 2\cos \left(\frac{4\theta +6\theta }{2}\right)\cos \left(\frac{4\theta -6\theta }{2}\right)=0$

$\Rightarrow 2\cos 5\theta \cos \theta =0$

$\Rightarrow \cos 5\theta =0$ or $\cos \theta =0$

$\Rightarrow 5\theta =(2n+1)\frac{\pi }{2}$ or $\theta =(2n+1)\frac{\pi }{2}$ for $n\in Z$

$\therefore \theta =(n+\frac{1}{2})\frac{\pi }{5}$ or $\theta =(n+\frac{1}{2})\pi ,n\in Z$