Question

If

$sec \: + tan \: = p \: \\ \\ then \: prove \: that \: \\ \\ \frac{ {p}^{2} - 1}{ {p}^{2} + 1} = sin \:$
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Answer 1

Answer:

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Answer 2

SOLUTION:-

Given:

$sec \theta \: + tan \theta = p$

To prove:

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

Proof:

$\frac{ {p}^{2} - 1}{ {p}^{2} + 1 } \\ \\ = > \frac{(sec \theta + tan \theta)2 - 1}{(sec \theta + tan \theta)2 + 1} \\ \\ = > \frac{sec2 \theta + tan2 \theta + 2sec \theta.tan \theta - 1}{sec2 \theta + tan 2\theta + 2sec \theta.tan \theta + 1} \\ \\ = > \frac{(sec2 \theta - 1) + tan2 \theta - 2sec \theta.tan \theta}{(tan2 \theta + 1) + sec2 \theta + 2sec \theta.tan \theta} \\ \\ = > \frac{tan2 \theta + tan2 \theta + 2sec \theta.tan \theta}{sec2 \theta + sec2 \theta + 2sec \theta.tan \theta} \\ \\ = > \frac{2tan \theta(tan \theta + sec \theta)}{2sec \theta(sec \theta + tan \theta)} \\ \\ = > \frac{tan \theta}{sec \theta} \\ \\ = > \frac{sin \theta}{cos \theta \times cos \theta} \\ \\ = > sin \theta$

R.H.S

[Proved]

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