Question
Prove that:
$
\sec \left(\frac{\pi}{4}+\theta\right) \sec \left(\frac{\pi}{4}-\theta\right)=2 \sec 2 \theta
$
$
\sec \left(\frac{\pi}{4}+\theta\right) \sec \left(\frac{\pi}{4}-\theta\right)=2 \sec 2 \theta
$
✓
Answer
$\begin{array}{l}\text { L.H.S. } \sec \left(\frac{\pi}{4}+\theta\right) \cdot \sec \left(\frac{\pi}{4}-\theta\right) \\ \Rightarrow \frac{1}{\cos \left(\frac{\pi}{4}+\theta\right) \cos \left(\frac{\pi}{4}-\theta\right)}=\frac{2}{2 \cos \left(\frac{\pi}{4}+\theta\right) \cos \left(\frac{\pi}{4}-\theta\right)} \\ \Rightarrow \frac{2}{\cos \left(\frac{\pi}{4}+\theta+\frac{\pi}{4}-\theta\right)+\cos \left(\frac{\pi}{4}+\theta-\frac{\pi}{4}+\theta\right)}\end{array}$
$[\because 2 \cos A \cos B=\cos (A+B)-\cos (A-B)]$
$\Rightarrow \frac{2}{\cos \frac{\pi}{2}+\cos 2 \theta}=\frac{2}{0+\cos 2 \theta}$
$\Rightarrow \frac{2}{\cos 2 \theta}=2 \sec 2 \theta=$ R.H.S.
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