Question

$Prove\:that\colon\\{(sinA+secA)}^{2}+\\{(cosA+cosecA)}^{2}\\={(1+secAcosecA)}^{2}$

Answer 1

Given,

$(\sin A+\sec A)^{2}+(\cos A+cosecA)^{2}\\ \\=\sin^{2}A+\sec^{2}A+2\sin A\sec A+\cos^{2}A+cosec^{2}A+2\cos A.cosecA\\\\=\sin^{2}A+cos^{2}A+\sec^{2}A+2\sin A\sec A+cosec^{2}A+2\cos A.cosecA\\\\=1+\sec^{2}A+cosec^{2}A+2\sin A.\sec A+2\cos A.cosecA\\\\=1+\frac{1}{\cos^{2}A}+\frac{1}{\sin^{2}A}+\frac{2\sin A}{\cos A}+\frac{2\cos A}{\sin A}\\\\=1+\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A\cos^{2}A}+2(\frac{\sin^{2}A+\cos^{2}A}{\sin A\cos A})\\\\=1+\frac{1}{\sin^{2}A\cos^{2}A}+2(\frac{1}{\sin A\cos A})\\\\=1+cosec^{2}A.\sec^{2}A+2\times1\times\sec A.cosecA\\ \\=(1+\sec A.cosecA)^{2}\\\\\textbf{Hence,Proved}$

$Note:\\1.\sin^{2}A+\cos^{2}A=1\\\\2.\sin A=\frac{1}{cosecA}\\\\3.\cos A=\frac{1}{\sec A}\\\\4.(a+b)^{2}=a^{2}+b^{2}+ab$

Answer 2

$\bold{\large{\underline{Solution\: \colon}}}$

$\bold{\underline{Revision\: of \: formula}}$

$\bold{\large{1) \: sec\: A = \dfrac{1}{cos\: A}}}$

$\bold{\large{2) \: cosec\: A = \dfrac{1}{sin\: A}}}$

$\bold{\large{3)\: sin^{2}\: A \: + \: cos^{2}\: A \: = \: 1}}$

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$\bold{\large{\underline{To\: Prove\: \colon}}}$


$\bold{\large{(sin\: A \: + sec \: A)^{2} \: + \: (cos\: A \: + \: cosec\: A)^{2}\: = \: (1\: + \: sec\: A \: .\:cosec\: A)^{2}}}$


$\bold{\large{\underline{Solution\: \colon}}}$


$\bold{\large{\underline{\underline{Take\: L.H.S.}}}}$



$\bold{\large{\therefore\:(sin\: A \: + \: sec\: A)^{2} + (cos \: A \: + \: cosec\: A)^{2}}}$


$\bold{\large{\implies \Bigg(sin\: A \: +\: \dfrac{1}{cos\: A}\Bigg)^{2} + \Bigg(cos\: A \: + \: \dfrac{1}{sin\: A}\Bigg)}}$


$\bold{\large{\implies \Bigg(\dfrac{sin\: A cos\: A \: - \: 1}{cos\: A}\Bigg)^{2} \: + \: \Bigg(\dfrac{sin\: A cos\: A \: - \: 1}{sin\: A}\Bigg)^{2}}}$


$\bold{\large{\implies \dfrac{(sin\: A cos\: A \: - \: 1)^{2} \: sin^{2}\: A + (sin\: A cos\: A \: -\: 1)^{2} \: cos^{2}\: A }{cos^{2}\: A sin^{2}\: A}}}$


$\bold{\large{\implies \dfrac{(sin\: A cos\: A\: + \: 1)^{2}[cos^{2}\: A \: + \: sin^{2}\: A]}{cos^{2}\: A \: + \: sin^{2}\: A}}}$


$\bold{\large{\implies \dfrac{(sin\: A cos\: A\: + \: 1)^{2}[1]}{cos^{2}\: A \: + \: sin^{2}\: A}}}$


$\bold{\large{\implies \dfrac{(sin\: A cos\: A\: + \: 1)^{2}}{cos^{2}\: A \: + \: sin^{2}\: A}}}$


$\bold{\large{\implies \dfrac{sin^{2}\: A cos^{2}\: A}{sin^{2}\: A \: cos^{2}\: A } \: + \: \dfrac{1}{sin^{2}\: A \: cos^{2}\: A} + \dfrac{2 \: sin\: A\: cos\: A}{sin^{2}\: A \: cos^{2}\: A}}}$


$\bold{\large{\implies 1 \: + \: cosec^{2}\: A . sec^{2}\: A \: + \: 2\: sec\: A \: cosec\: A }}$


$\bold{\large{\implies 1 \: + \: cosec\: A sec\: A \: + \: cosec\: A \: sec\: A \: +\:cosec^{2}\: A . sec^{2}\: A \: }}$


$\bold{\large{\implies [1(1\: + \: sec\: A \: cosec\: A ) + sec\: A \: cosec\: A( 1\: +\:sec\: A \: cosec\: A )]}}$


$\bold{\large{\implies (1\: + \: sec\: A \: cosec\: A)(1\: + \: sec\: A \: cosec\: A)}}$


$\bold{\large{\implies (1\:+\: sec\: A \: cosec\: A )^{2} }}$


$\bold{\underline{\underline{R.H.S}}}$

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$\bold{\large{\boxed{LHS\: = \: RHS }}}$

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$\bold{\large{Thanks}}$