Question

this is my question someone solve this​

Answer 1

hope it helps you. thanks for asking

Answer 2

$\huge{\underline{\underline{\green{\mathsf{Answer:}}}}}$

Given

$\mathsf{cosecθ + cotθ = p \rightarrow (1)}$

$\mathsf{Using \:the\: trigonometric\: identity}$

$\fbox{\mathsf{cosec^2\theta = 1 + cot^2\theta}}$

$\mathsf{\implies\:cosec^2θ - cot^θ = 1}$

Using the identity

$\fbox{\mathsf{a^2 - b^2 = (a+b) (a-b) }}$

$\mathsf{\implies\: (cosecθ + cotθ) (cosecθ - cotθ) = 1}$

$\mathsf{\implies\:p ( cosecθ - cotθ ) = 1}$

$\mathsf{\implies\: cosecθ - cotθ = \frac{1}{p}\rightarrow(2) }$

Adding (1) and (2) we get

cosecθ + cotθ = p

cosecθ – cotθ = 1/p

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$\mathsf{\implies\: 2cosecθ \: \: \:\:\:\:\:\: = p + \frac{1}{p}}$

$\mathsf{\implies\: 2cosecθ \: \: \:\:\:\:\:\: = \frac{p^2 + 1}{p}}$

$\mathsf{\implies\: cosec\theta \: \: \:\:\:\:\:\: = \frac{p^2 + 1}{2p}}$

$\fbox{\mathsf{We\:know\:that,\:\:cosec\theta = \frac{1}{sin\theta}}}$

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

$\fbox{\mathsf{We\:know\:that,\:\:cos\theta = \sqrt{1 - sin^2\theta}}}$

$\mathsf{\implies\: cos\theta \: \: \:\:\:\:\:\: = \sqrt{1 - (\frac{2p}{p^2 + 1})^2}}$

$\mathsf{\implies\: cos\theta \: \: \:\:\:\:\:\: = \sqrt{\frac{p^4 + 1 + 2p^2 - 4p^2}{(p^2 + 1)^2}}}$

$\mathsf{\implies\: cos\theta \: \: \:\:\:\:\:\: = \sqrt{\frac{p^4 + 1 - 2p^2}{(p^2 + 1)^2}}}$

$\mathsf{\implies\: cos\theta \: \: \:\:\:\:\:\: = \sqrt{\frac{(p^2 - 1)^2}{(p^2 + 1)^2}}}$

$\fbox{\mathsf{\implies\: cos\theta \: \: \:\:\:\:\:\: = \frac{p^2 - 1}{p^2 + 1}}}$

Hence Proved