 ## Book: RD Sharma - Mathematics

### Chapter: 7. Values of Trigonometric Functions at Sum of Difference of Angles

#### Subject: - Class 11th

##### Q. No. 16 of Exercise 7.1

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16
##### Prove that:cos2A + cos2B – 2 cosA cosB cos(A +B) = sin2(A +B)

LHS = cos2A + cos2B – 2 cosA cosB cos(A +B)

= cos2A + 1 – sin2B - 2 cosA cosB cos(A +B)

= 1 + cos2A – sin2B - 2 cosA cosB cos(A +B)

We know that cos2A – sin2B = cos(A +B) cos(A –B)

= 1 + cos(A +B) cos(A –B) - 2 cosA cosB cos(A +B)

= 1 + cos(A +B) [cos(A –B) – 2 cosA cosB]

We know that cos(A -B) = cosA cosB + sinA sinB.

= 1 + cos(A +B) [cosA cosB + sinA sinB – 2 cosA cosB]

= 1 + cos(A +B) [-cosA cosB + sinA sinB]

= 1 - cos(A +B) [cosA cosB - sinA sinB]

We know that cos(A +B) = cosA cosB - sinA sinB.

= 1 – cos2(A +B)

= sin2(A +B) = RHS

Hence proved.

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