prove that: ^[1-sin^2A(90-A)] + ^[4-4sin^2(90-A)] = 3sinA
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Hi ,
[1 - sin² ( 90 - A )] + [ 4 - 4sin²(90 - A )]
= 3sinA
LHS =[1-sin²(90-A)]+[4-4sin²(90-A)]
= [ 1 - cos² A ] + [ 4 - 4cos²A ]
= sin² A + 4( 1 - cos² A )
= Sin² A + 4sin² A
= 5sin²A
Plz , check the answer .
I hope this helps you.
: )
[1 - sin² ( 90 - A )] + [ 4 - 4sin²(90 - A )]
= 3sinA
LHS =[1-sin²(90-A)]+[4-4sin²(90-A)]
= [ 1 - cos² A ] + [ 4 - 4cos²A ]
= sin² A + 4( 1 - cos² A )
= Sin² A + 4sin² A
= 5sin²A
Plz , check the answer .
I hope this helps you.
: )
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