Question

Q9.
Prove that : sin^2(n + 1) A – sin^2 nA = sin(2n + 1) A sin A
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Answer 1

Answer:

Step-by-step explanation:

We know that sin2A – sin2B = sin(A +B) sin(A –B)

HereA =(n + 1)A And B = nA

⇒ LHS: sin2(n + 1)A – sin2nA = sin((n + 1)A + nA) sin((n + 1)A – nA)

= sin(nA +A + nA) sin(nA +A – nA)

= sin(2nA +A) sin(A)

= sin(2n + 1)A sinA = RHS

Hence proved

Answer 2

Answer:

sin2(n+1)A - sin2nA = sin(2n+1)A.sinA

we have formula sin2a - sin2b =sin(a-b)sin(a+b)

therefore

sin2(n+1)A - sin2nA=sin[(n+1)A+nA].sin[(n+1)A-nA]

=sin(2n+1)A.sinA =RHS

hence proved