Question

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


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Answer 1

$sec\theta + tan\theta = p$

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

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

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

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

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

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

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

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

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

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

$\implies sin\theta$

Thus

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

Hence proved

Hope it helps you :-)

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Answer 2

Heya !!


Ur answer ⤵️⤵️

secθ+tanθ=p

p^2 = (sec\theta + \frac{sin\theta}{cos\theta})^2p2=(secθ+cosθsinθ​)2

\implies p^2=\frac{(1+sin\theta)^2}{cos^2\theta}⟹p2=cos2θ(1+sinθ)2​

\implies p^2 =\frac{(1+sin\theta)^2}{1-sin^2\theta}⟹p2=1−sin2θ(1+sinθ)2​

\implies p^2=\frac{(1+sin\theta)^2}{(1+sin\theta)(1-sin\theta)}⟹p2=(1+sinθ)(1−sinθ)(1+sinθ)2​

\implies p^2=\frac{1+sin\theta}{1-sin\theta}⟹p2=1−sinθ1+sinθ​

p^2-1 =\frac{1+sin\theta}{1-sin\theta}-1p2−1=1−sinθ1+sinθ​−1

\implies p^2-1=\frac{1+sin\theta-1+sin\theta}{1-sin\theta}⟹p2−1=1−sinθ1+sinθ−1+sinθ​

\implies p^2-1=\frac{2sin\theta}{1-sin\theta}⟹p2−1=1−sinθ2sinθ​ .................(1)

p^2+1=\frac{1+sin\theta+1-sin\theta}{1-sin\theta}[p2+1=1−sinθ1+sinθ+1−sinθ​[

p^2+1=\frac{2}{1+sin\theta}p2+1=1+sinθ2​ ........................(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}}p2+1p2−1​=1−sinθ2​1−sinθ2sinθ​​

\implies \frac{2 sin\theta}{2}⟹22sinθ​

\implies sin\theta⟹sinθ

Thus

\boxed{\frac{p^2-1}{p^2+1}=sin\theta}p2+1p2−1​=sinθ​


Hope that helps ❣❣


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