Question

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

Answer 1

Answer:

Hi ,

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

I hope this helps you.

:)

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Helping Hand

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9 people helped

hi,

LHS

=

[(1+ sinA - cosA) /

(1+ sinA + cosA)]²

according to identity ,

(a+b-c)² = a²+b²+c²+2ab-2bc-2ac

=( 1 + sin²A + cos²A + 2sinA - 2sinA.cosA - 2cosA ) /

( 1 + sin²A + cos²A + 2sinA + 2sinA.cosA + 2cosA)

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

( 1 + 1 + 2sinA + 2cosA + 2sinA.cosA)

=( 2 + 2sinA - 2cosA - 2sinA.cosA ) /

( 2 + 2sinA + 2cosA + 2sinA.cosA)

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

2( 1 + sinA + cosA + sinA.cosA)

=( 1 + sinA - cosA - sinA.cosA ) /

( 1 + sinA + cosA + sinA.cosA)

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

[ 1(1 + sinA) +cosA(1 + sinA)]

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

[ (1 + cosA)(1 + sinA)]

=( 1 - cosA ) / = RHS ,hence proved

( 1 + cosA)