Question

If sec $\theta$ + tan $\theta =p,$ then $\sin \theta =$

• A. $\frac{p^2+1}{p^2-1}$
• B. $\frac{p^2-1}{p^2+1}$
• C. $\frac{1-p^2}{1+p^2}$
• D. $\frac{1+p^2}{1-p^2}$

Answer 1

The correct option is B $\frac{p^2-1}{p^2+1}$

Given $\sec \theta +\tan \theta =p$ $\dots \left(I\right)$

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

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

$\Rightarrow \left(p\right)(\sec \theta -\tan \theta )=1$

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

Add $\left(I\right)$ and $\left(II\right)$

we get $2\sec \theta =p+\frac{1}{p}$

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

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

Subtract $\left(I\right)$ and $\left(II\right)$

we get $2\tan \theta =p-\frac{1}{p}$

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

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

divide $\tan \theta$ and $\sec \theta$

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

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

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