If + =p, prove that: = p^2-1 p^2+1

Question

If $\sec\theta+\tan\theta=\text{p},$ prove that:
$\sin\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$

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Answer

We have,
$\sec\theta+\tan\theta=\text{p}\dots(1)$
$\Rightarrow\frac{\sec\theta+\tan\theta}{1}\times\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{1}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\sec\theta-\tan\theta=\frac{1}{\text{p}}\dots(2)$
Adding $(1)$ and $(2)$,, we get:
$2\sec\theta=\Big(\text{p}+\frac{1}{\text{p}}\Big)$
$\Rightarrow\sec\theta=\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)\dots(\text{A})$
Subtracting $(2)$ and $(1)$,, we get:
$2\tan\theta=\Big(\text{p}-\frac{1}{\text{p}}\Big)$
$\Rightarrow\tan\theta=\frac12\Big(\text{p}-\frac{1}{\text{p}}\Big)\dots(\text{B})$
Using $(A)$ and $(B)$,, we get:
$\sin\theta=\frac{\tan\theta}{\sec\theta}$
$=\frac{\frac12\Big(\text{p}-\frac{1}{\text{p}}\Big)}{\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)}$
$=\frac{\Big(\frac{\text{p}^2-1}{\text{p}}\Big)}{\Big(\frac{\text{p}^2+1}{\text{p}}\Big)}$
$\therefore\ \sin\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$

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