If (sec θ + tan θ ) = p then show that (sec θ – tan θ)

Given: $\sec \theta+\tan \theta=p \quad \ldots$ (1)

We know

$\sec ^{2} \theta-\tan ^{2} \theta=1$

$\Rightarrow(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)=1$

$\Rightarrow(\sec \theta-\tan \theta) p=1 \quad[$ From $(1)]$

$\Rightarrow \sec \theta-\tan \theta=\frac{1}{p} \quad \ldots \ldots(2)$

Adding (1) and (2), we get

$\sec \theta+\tan \theta+\sec \theta-\tan \theta=p+\frac{1}{p}$

$\Rightarrow 2 \sec \theta=\frac{p^{2}+1}{p}$

$\Rightarrow \sec \theta=\frac{p^{2}+1}{2 p}$

$\Rightarrow \frac{1}{\sec \theta}=\frac{2 p}{p^{2}+1}$

$\Rightarrow \cos \theta=\frac{2 p}{p^{2}+1} \quad \ldots \ldots(3)$

Subtracting (2) from (1), we get

$\sec \theta+\tan \theta-\sec \theta+\tan \theta=p-\frac{1}{p}$

$\Rightarrow 2 \tan \theta=\frac{p^{2}-1}{p}$

$\Rightarrow \tan \theta=\frac{p^{2}-1}{2 p} \quad \ldots \ldots(4)$

Now,

$\sin \theta=\tan \theta \times \cos \theta$

$\Rightarrow \sin \theta=\frac{p^{2}-1}{2 p} \times \frac{2 p}{p^{2}+1} \quad[$ Using $(3)$ and $(4)]$+

$\Rightarrow \sin \theta=\frac{p^{2}-1}{p^{2}+1}$