Question

If (sec θ + tan θ ) = p then show that (sec θ – tan θ) = $\frac{1}{p}$.
Hence, show that cos θ = $\frac{2p}{\left(p^2+1\right)}\mathrm{and}\sin \mathrm{\theta }=\frac{p^2-1}{p^2+1}$.

Answer 1

Given: $\sec \theta +\tan \theta =p.....\left(1\right)$
We know
$$\begin{gathered} {\sec }^2\theta -{\tan }^2\theta =1 \\ \Rightarrow \left(\sec \theta -\tan \theta \right)\left(\sec \theta +\tan \theta \right)=1 \\ \Rightarrow \left(\sec \theta -\tan \theta \right)p=1\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow \sec \theta -\tan \theta =\frac{1}{p}.....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$$\begin{gathered} \sec \theta +\tan \theta +\sec \theta -\tan \theta =p+\frac{1}{p} \\ \Rightarrow 2\sec \theta =\frac{p^2+1}{p} \\ \Rightarrow \sec \theta =\frac{p^2+1}{2p} \\ \Rightarrow \frac{1}{\sec \theta }=\frac{2p}{p^2+1} \end{gathered}$$
$\Rightarrow \cos \theta =\frac{2p}{p^2+1}.....\left(3\right)$
Subtracting (2) from (1), we get
$$\begin{gathered} \sec \theta +\tan \theta -\sec \theta +\tan \theta =p-\frac{1}{p} \\ \Rightarrow 2\tan \theta =\frac{p^2-1}{p} \\ \Rightarrow \tan \theta =\frac{p^2-1}{2p}.....\left(4\right) \end{gathered}$$
Now,
$$\begin{gathered} \sin \theta =\tan \theta \times \cos \theta \\ \Rightarrow \sin \theta =\frac{p^2-1}{2p}\times \frac{2p}{p^2+1}\left[\mathrm{Using}\left(3\right)\mathrm{and}\left(4\right)\right] \\ \Rightarrow \sin \theta =\frac{p^2-1}{p^2+1} \end{gathered}$$