Question

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Ques:-. If sin theta + cos theta = p, Prove that sin theta =
$\frac{ {p}^{2} - 1}{ {p}^{2} + 1}$


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Answer 1

Given:

$sin\theta + cos\theta = p$

$p^2 = (sin\theta + cos\theta)^2$

$\implies p^2 = sin^2\theta + cos^2\theta + 2sin\theta cos\theta$

$\implies p^2 = 1 + 2 sin\theta cos\theta$................(1)

$p^2 - 1 = 2sin\theta cos\theta$.........................(2)

$p^2 + 1 = 2 + 2sin\theta cos\theta$

Hey I guess your question is wrong

The correct question will be:

$sec\theta + tan\theta = p$

$p^2 = (sec\theta + \frac{sin\theta}{cos\theta})^2$

$\implies p^2=\frac{(1+sin\theta)^2}{cos^2\theta}$

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

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

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

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

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

$\implies p^2-1=\frac{2sin\theta}{1-sin\theta}$.................(1)

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

$p^2+1=\frac{2}{1+sin\theta}$........................(2)

Dividing 1 and 2

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

$\implies \frac{2 sin\theta}{2}$

$\implies sin\theta$

So:

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

Hence proved

Hope it helps you

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Answer 2

hey mate

here's the solution