If $(\sec \theta-\tan \theta)=\sqrt{2} \tan \theta$ then prove that $(\sec \theta+\tan \theta)=\sqrt{2} \sec \theta$.
$\sec \theta-\tan \theta=\sqrt{2} \tan \theta$
Squaring on both sides, we get
$(\sec \theta-\tan \theta)^{2}=(\sqrt{2} \tan \theta)^{2}$
$\Rightarrow \sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta=2 \tan ^{2} \theta$
$\Rightarrow \sec ^{2} \theta-\tan ^{2} \theta=2 \sec \theta \tan \theta$
$\Rightarrow(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)=2 \sec \theta \tan \theta \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$
$\Rightarrow \sqrt{2} \tan \theta(\sec \theta+\tan \theta)=2 \sec \theta \tan \theta \quad(\sec \theta-\tan \theta=\sqrt{2} \tan \theta)$
$\Rightarrow \sec \theta+\tan \theta=\frac{2 \sec \theta \tan \theta}{\sqrt{2} \tan \theta}$
$\Rightarrow \sec \theta+\tan \theta=\sqrt{2} \sec \theta$
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