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(1+sin2A+cos2A)^2=4cos^2A(1+sin2A)
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(1+sin2A+cos2A)^2=4cos^2A(1+sin2A)

User Justin Leo
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1 Answer

4 votes

Answer:

(1+sin2A+cos2A) ^2=4cos^2A(1+sin2A)

Explanation:

Here,

L.H.S=(1+sin2A+cos2A)^2

=(1+2sinAcosA+2cos^2A-1)^2

=(2sinAcosA+2cos^2A)^2

={2cosA(sinA+cosA)}^2

=4cos^2A(sinA+cosA)^2

=4cos^2A(sin^2A+2sinA.cosA+cos^2A)

=4cos^2A(sin^2A+cos^A+2sinAcosA)

=4cos^2A(1+sin2A)

=R.H.S

Hence,L.H.S=R.H.S.

(1+sin2A+cos2A)^2=4cos^2A(1+sin2A).

proved.

User Jordonias
by
3.4k points
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