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(1+sinA-cosA/1+sinA+cosA)^2 = 1-cosA/1+cosA

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(1+sinA-cosA/1+sinA+cosA)^2 = 1-cosA/1+cosA

Posted by Anushka Gurjar 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 10 months ago

LHS = [(1+sinA-cosA )/(1+sinA+cosA)]²

=[ (1+sin²A+cos²A+2sinA-2sinAcosA-2cosA)/(1+sin²A+cos²A+2sinA+2sinAcosA+2cosA) ]²

= [( 1+1+2sinA-2sinAcosA-2cosA)/(1+1+2sinA+2sinAcosA+2cosA)]²

= { ( 2(1+sinA) -2cosA(sinA+1) / [ 2(1+sinA) + 2cosA(sinA +1 ) ]}²

= { (2(1+sinA)[1 - cosA] /[2(1+sinA )(1 + cosA )] }²

= [ ( 1 - cosA ) / ( 1 + cosA ) ]²

= RHS

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