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Sin(n+2)A.cos(n+1)A-cos(n+2)A.sin(n+1)A=sinA​

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5 votes

Answer:

Explanation:

Instead of theeta let us take A.

LHS =(T3 - T5)/T1

= {(sin^3 A+cos^3 A) - (sin^5 A+cos^5 A) /(sin A+ cos A)

= (sin^3 A-sin^5 A +cos^3 A-cos^5 A) / (sin A + cosA)

=sin^3 A(1-sin^2A) + cos^3 A(1-cos^2 A) / (sinA+cosA)

= (sin^3 Acos^2 A + cos^3 A.sin^2 A) / (sinA+cosA)

=sin^2A.cos^2A(sinA+cosA)/(sinA+cosA)

= sin^2 A.cos^2 A. ……..(1)

RHS = (T5-T7)/T3 =

{(sin^5 A+cos^5 A) - (sin^7 A+ cos^7 A) / (sin^3 A+ cos^3 A)

=(sin^5 A - sin^7 A + cos^5 A - cos^7 A) / (sin^3 A + cos^3 A)

={sin^5 A(1-sin^2 A)+cos^5 A(1-cos^2 A)} / (sin^3 A + cos^3 A)

=(sin^5 A.cos^2 A + cos^5 A.sin^2 A) / (sin^3 A + cos^3 A)

=sin^2 A.cos^2 A(sin^3 A + cos^3 A) / (sin^3 A + cos^3 A)

= sin^2 A.cos^2 A ….(2)

From (1) and (2) LHS = RHS

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