Question

If $\sec \theta +\tan \theta =p$ then $\sin \theta =\frac{p^2+1}{p^2-1}$

• A. True
• B. False

Answer 1

The correct option is B False

$\sec \theta +\tan \theta =p$ .........$\left(1\right)$

$(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=p(\sec \theta -\tan \theta )$

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

$\Rightarrow 1=p(\sec \theta -\tan \theta )$ since ${\sec }^2\theta -{\tan }^2\theta =1$

$\Rightarrow \sec \theta -\tan \theta =\frac{1}{p}$ ......$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

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

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

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

Equations $\left(1\right)-\left(2\right)$ we get

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

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

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

Now $\frac{\tan \theta }{\sec \theta }=\frac{\frac{p^2-1}{2p}}{\frac{p^2+1}{2p}}=\frac{p^2-1}{p^2+1}$

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

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