Question

sin3A+sinA/cos3A+cosA​

Answer 1

Answer:

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Answer 2

Answer:

LHS

Sin3A/SinA-Cos3A/CosA

=(Sin3A)(CosA) -(Cos3A)(SinA)/SinACosA

=(3SinA - 4sin3A)CosA-(4cos3A-3CosA)(SinA)/sinAcosA

=3 SinAcosA -4sin3 cosA-4cos3AsinA+3 sin AcosA/ SinACosA

=6SinAcosA-4Sin3cosA-4cos3SijA/sinAcosA

=SinAcosA(6-4sin2A-4cos2A)/sinAcosA

=6-4Sin2A - 4cis2A

6 - 4(Sin2A+cis2A)[Sin2A+ cos2h= 1

= 6-4×1

=6-4

= 2

RHS

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