Question

If $\sec \theta +\tan \theta =p$, prove that

$\left(\mathrm{i}\right)\sec \theta =\frac{1}{2}\left(p+\frac{1}{p}\right)\left(\mathrm{ii}\right)\tan \theta =\frac{1}{2}\left(p-\frac{1}{p}\right)\left(\mathrm{iii}\right)\sin \theta =\frac{p^2-1}{p^2+1}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{We}\mathrm{have}, \\ \sec \theta +\tan \theta =p.....\left(1\right) \\ \Rightarrow \frac{\sec \theta +\tan \theta }{1}\times \frac{\sec \theta -\tan \theta }{\sec \theta -\tan \theta }=p \\ \Rightarrow \frac{{\sec }^2\theta -{\tan }^2\theta }{\sec \theta -\tan \theta }=p \\ \Rightarrow \frac{1}{\sec \theta -\tan \theta }=p \\ \Rightarrow \sec \theta -\tan \theta =\frac{1}{p}.....\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2\sec \theta =p+\frac{1}{p} \\ \Rightarrow \sec \theta =\frac{1}{2}\left(p+\frac{1}{p}\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Subtracting}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get} \\ 2\tan \theta =\left(p-\frac{1}{p}\right) \\ \Rightarrow \tan \theta =\frac{1}{2}\left(p-\frac{1}{p}\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Using}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get} \\ \sin \theta =\frac{\tan \theta }{sec\theta } \\ =\frac{{\displaystyle \frac{1}{2}}\left(p-{\displaystyle \frac{1}{p}}\right)}{{\displaystyle \frac{1}{2}}\left(p+{\displaystyle \frac{1}{p}}\right)} \\ =\frac{\left({\displaystyle \frac{p^2-1}{p}}\right)}{\left({\displaystyle \frac{p^2+1}{p}}\right)} \\ \therefore \sin \theta =\frac{p^2-1}{p^2+1} \end{gathered}$$