Question

If $\tt \: p \: = \: sin\theta \: + \: cos\theta$
$\tt \: q \: = \: cosec\theta \: + \: sec\theta$

Then prove :-

$\tt\: q( {p}^{2} - 1) \: = 2p$​

Answer 1

Identities Used :-

$\: \begin{gathered}\:{\underline{\boxed{\bf{\red{{\sf \: \: sec\theta \: = \: \dfrac{1}{cos \: \theta} }}}}}} \\ \end{gathered}$

$\: \begin{gathered}\:{\underline{\boxed{\bf{\purple{{\sf \: \: cosec\theta \: = \: \dfrac{1}{sin \: \theta} }}}}}} \\ \end{gathered}$

$\: \begin{gathered}\:{\underline{\boxed{\bf{\green{{\sf \: \: {sin}^{2}\theta + {cos}^{2}\theta = 1 }}}}}} \\ \end{gathered}$

$\: \begin{gathered}\:{\underline{\boxed{\bf{\purple{{\sf \: \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }}}}}} \\ \end{gathered}$

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Given :-

• $\sf \: p \: = \: sin\theta \: + \: cos\theta$
• $\sf \: q \: = \: cosec\theta \: + \: sec\theta$

To Prove :-

$\sf\: q( {p}^{2} - 1) \: = 2p$

Proof:-

$= \bf \: LHS$

$\rm :\implies\:q( {p}^{2} - 1)$

$\rm :\implies\:(sec\theta + cosec\theta)( {(sin\theta + cos\theta)}^{2} - 1)$

$\rm :\implies\: \bigg(\dfrac{1}{cos\theta} + \dfrac{1}{sin\theta} \bigg)\bigg( {sin}^{2}\theta + {cos}^{2}\theta + 2sin\theta \: cos\theta - 1) \bigg)$

$\rm :\implies\:\bigg( \dfrac{sin\theta + cos\theta}{sin\theta \: cos\theta} \bigg)\bigg( 1 + 2sin\theta \: cos\theta \: - 1\bigg)$

$\rm :\implies\:\dfrac{p}{sin\theta \: cos\theta} \times 2 \: sin\theta \: cos\theta$

$\rm :\implies\:2 \: p$

$= \bf \: RHS$

• Hence,(Proved..!)

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