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Prove that LHS=RHS

$\frac{1}{ \cos \: A + \sin A - 1} + \frac{1}{ \cos \:A + \sin \: A +1 }$
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Answer 1

Answer:

LHS.

$\\ \longrightarrow\ \sf\frac{1}{ \cos \: A + \sin A - 1} + \frac{1}{ \cos \:A + \sin \: A +1 }\\ \\ \longrightarrow\sf \frac{ \cos \: A + \sin \:A + 1 \cos \: A + \sin \: A }{( \cos \: A + \sin \: A) {}^{2} - 1}$

$\\ \\ \sf \longrightarrow\frac{2(cosA+sinA)}{sos^2+sin^2 A +2cos A sinA-1}\\ \\ \sf\longrightarrow \frac{2( \cos \:A + \sin \:A) }{1 + 2 \cos \: A \sin \: A } = \frac{ \cos \: A + \sin \: A }{ \cos \:A \sin \: A}\\ \\ \sf\longrightarrow \sf \frac{ \cos \: A}{ \cos \: A \sin \: A} + \frac{ \sin \: A }{ \cos \: A \: \sin \: A }\\ \\ \sf\longrightarrow \frac{1}{ \sin \: A } + \frac{1}{ \cos \: A}\\ \\ \sf\longrightarrow cosecA + secA= RHS.$

• LHS=RHS

Answer 2

Step-by-step explanation:

$LHS.</p><p>\begin{gathered}\\ \longrightarrow\ \sf\frac{1}{ \cos \: A + \sin A - 1} + \frac{1}{ \cos \:A + \sin \: A +1 }\\ \\ \longrightarrow\sf \frac{ \cos \: A + \sin \:A + 1 \cos \: A + \sin \: A }{( \cos \: A + \sin \: A) {}^{2} - 1}\end{gathered}⟶ cosA+sinA−11+cosA+sinA+11⟶(cosA+sinA)2−1cosA+sinA+1cosA+sinA</p><p>\begin{gathered} \\ \\ \sf \longrightarrow\frac{2(cosA+sinA)}{sos^2+sin^2 A +2cos A sinA-1}\\ \\ \sf\longrightarrow \frac{2( \cos \:A + \sin \:A) }{1 + 2 \cos \: A \sin \: A } = \frac{ \cos \: A + \sin \: A }{ \cos \:A \sin \: A}\\ \\ \sf\longrightarrow \sf \frac{ \cos \: A}{ \cos \: A \sin \: A} + \frac{ \sin \: A }{ \cos \: A \: \sin \: A }\\ \\ \sf\longrightarrow \frac{1}{ \sin \: A } + \frac{1}{ \cos \: A}\\ \\ \sf\longrightarrow cosecA + secA= RHS.\end{gathered}⟶sos2+sin2A+2cosAsinA−12(cosA+sinA)⟶1+2cosAsinA2(cosA+sinA)=cosAsinAcosA+sinA⟶cosAsinAcosA+cosAsinAsinA⟶sinA1+cosA1⟶cosecA+secA=RHS.</p><p></p><p>LHS=RHS</p><p></p><p>$