Question

Prove that:
$\sec \left(\frac{\pi }{4}+\theta \right)\sec \left(\frac{\pi }{4}-\theta \right)=2\sec 2\theta$

Answer 1

$\begin{array}{c}LHS=\sec \left(\frac{\pi }{4}+\theta \right)\sec \left(\frac{\pi }{4}-\theta \right)=\frac{1}{\cos \left(\frac{\pi }{4}-\theta \right)}\times \frac{1}{\cos \left(\frac{\pi }{4}+\theta \right)}\\ \frac{2}{2\cos \left(\frac{\pi }{4}-\theta \right)\cos \left(\frac{\pi }{4}+\theta \right)}=\frac{2}{\cos \left(\frac{2\pi }{4}\right)+\cos \left(2\theta \right)}\\ =\frac{2}{\cos \left(\frac{\pi }{2}\right)+\cos \left(2\theta \right)}=\frac{2}{0+\cos \left(2\theta \right)}=\frac{2}{\cos \left(2\theta \right)}=2\sec 2\theta =RHS\end{array}$

Hence L.H.S=R.H.S