If (sec θ – tan θ)

$\sec \theta-\tan \theta=\sqrt{2} \tan \theta$

Squaring on both sides, we get

$(\sec \theta-\tan \theta)^{2}=(\sqrt{2} \tan \theta)^{2}$

$\Rightarrow \sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta=2 \tan ^{2} \theta$

$\Rightarrow \sec ^{2} \theta-\tan ^{2} \theta=2 \sec \theta \tan \theta$

$\Rightarrow(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)=2 \sec \theta \tan \theta \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$

$\Rightarrow \sqrt{2} \tan \theta(\sec \theta+\tan \theta)=2 \sec \theta \tan \theta \quad(\sec \theta-\tan \theta=\sqrt{2} \tan \theta)$

$\Rightarrow \sec \theta+\tan \theta=\frac{2 \sec \theta \tan \theta}{\sqrt{2} \tan \theta}$

$\Rightarrow \sec \theta+\tan \theta=\sqrt{2} \sec \theta$