Question

If sin theta + cos theta is equal to p and sec theta + cosec theta is equal to q then prove that q into p square minus 1 is equal to p

Answer 1

$\huge{\underline{\underline{\blue{\mathfrak{Answer :}}}}}$

$\huge{\underline{\textbf{Given \: :}}}$

P = Sinθ + Cosθ

Q = Secθ + Cosecθ

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$\huge{\underline{\bf{To \: Prove \: :}}}$

Q(P² - 1) = 2P

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$\huge{\underline{\bf{Proof\: :}}}$

Q(P² - 1) = 2P ......... (1)

Put Value of P and Q in equation 1.

$\sf{(sec \: \theta \: + \: cosec \: \theta) (sin \:\theta \: + {cos \: \theta)}^{2} - 1 } \\ \\ \large{\bf{Using \: identity}} \\ \\\large{\boxed{\boxed{\purple{\mathfrak{(a \: + \: b {)}^{2} = {a}^{2} + {b}^{2} + 2ab }}}}} \\ \\ \sf{( \frac{1}{cos \: \theta} } + \frac{1}{sin \:\theta } )( {sin}^{2} \theta \: + \: {cos}^{2} \theta \: + 2sin \: \theta \: cos \: \theta - 1) \\ \\ \large{\bf{Using \: idenity: }} \\ \\ \large{\boxed{\boxed{\red{\mathfrak{ {sin}^{2}\theta \: + {cos}^{2} \theta = 1 }}}}} \\ \\ \sf{ (\frac{sin \: \theta \: + \: cos \theta}{sin\theta \: cos\theta} })( \cancel1 - \cancel1 + 2sin \: \theta \: cos \: \theta) \\ \\ \sf{(\frac{sin \: \theta \: + \: cos \theta}{ \cancel{sin\theta \: cos\theta}})( \cancel{2sin \: \theta \: cos \: \theta)}} \\ \\ \sf{2(sin \:\theta \: + cos \: \theta )} \\ \\ \sf{2P}$

$\huge{\boxed{\boxed{\pink{\mathcal{ Hence\: Proved}}}}}$

Answer 2

Answer:

★ Refer to attachment for solution.★

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