Math, asked by moon1237879, 1 year ago

Prove that :-
{(1+sinA-cos A) / (1+sinA+cos A)}^2=(1-cosA) / (1+cos A)

Answers

Answered by mysticd
198
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.

:)
Answered by rajindersingh3pcmzfe
161
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)

Thank You☺️
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