Math, asked by harinakhati, 3 months ago

prove that :cos^2 A + sin^2 A.cos 2B = cos^2 B + sin^2 ß.cos2A​

Answers

Answered by sijajajajajaja
1

Step-by-step explanation:

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Answered by barani7953
1

Step-by-step explanation:

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L.H.S=cos

2

A+cos

2

B−2cosAcosBcos(A+B)

=cos

2

A+cos

2

B−2cosAcosB(cosAcosB−sinAsinB)

=cos

2

A+cos

2

B−2cos

2

Acos

2

B+2cosAsinAcosBsinB

R.H.S=sin

2

(A+B)

=(sin(A+B))

2

=(sinAcosB+cosAsinB)

2

=sin

2

Acos

2

B+cos

2

Asin

2

B+2sinAcosAsinBcosB

=(1−cos

2

A)cos

2

B+cos

2

A−cos

2

Acos

2

B

=cos

2

B−cos

2

A+cos

2

B+cos

2

A−cos

2

Acos

2

B.

=cos

2

A+cos

2

B−2cos

2

Acos

2

B+2cosAsinAcosBSinB

Hence,

L.H.S=R.H.S

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