Question

(1+sinA-COSA)
(1+sinA+COSA)=(1-COSA)(1+COSA)^2​

Answer 1

Answer:

Step-by-step explanation:

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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Answer 2

Answer:

hiii...

how are you??

here is your 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 it will help u...

please please mark as brainleast please please...

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