Question

Cos^2(A-B)+cos^2B-2cos(A-B)cosAcosB=sin^2A

Answer 1

Answer:

proved

Step-by-step explanation:

Given Cos^2(A-B)+cos^2 B-2 cos(A-B) cos A cos B=sin^2 A

We know that

cos(A - B) = cos A cos B + sin A sin B and (a - b) ^2 = a^2 + 2 a b + b^2

Applying this formula we get

(cos A cos B + sin A sin B)^2 + cos^2 B - 2(cos A cos B + sin A sin B)cos A cos B

cos^2 A cos^2 B + 2 cos A cos B sin A sin B + sin^2 A sin^2 B + cos^2 B - 2 cos^2 A cos^2 B - 2 sin A sin B cos A cos B

cos^2 A cos^2 B + sin^2 A sin^2 B + cos^2 B - 2 cos^2 A cos^2 B

sin^2 A sin^2 B - cos^2 A cos^2 B + cos^2 B

cos^2 B - cos^2 A cos^2 B + sin^2 A sin^2 B

cos^2 B(1 - cos^2 A) + sin^2 A sin^2 B

cos^2 B sin^2 A + sin^2 A sin^2 B

sin^2 A(cos^2 B + sin^2 B)

sin^2 A(1)

sin^2 A

Answer 2

Answer:

${ \sin }^{2} a$

Step-by-step explanation:

${cos}^{2} (a - b) + {cos}^{2} \: \: b \: - 2 \cos(a + b) \cos(a) \cos(b) \\ \\ = {( \cos(a) \cos(b) + \sin(a) \sin(b))}^{2} + {cos}^{2} b \: - 2( \cos(a) \cos(b) + \sin(a) \sin(b) ) \cos(a) \cos(b) \\ \\ = {cos}^{2} a \: \: {cos}^{2} b + 2 \cos(a) \cos(b) \sin(a) \sin(b) + {sin}^{2} (a) {sin}^{2} (b) + {cos}^{2} (b) - 2( {cos}^{2} (a) {cos}^{2} (b) \\ \\ = {cos}^{2} b - {cos}^{2} a \: \: {cos}^{2} b \: + {sin}^{2} a \: \: {sin}^{2} b \\ \\ = {cos}^{2} b(1 - {cos}^{2} a) + {sin}^{2} a \times {sin}^{2} b \\ \\ = {cos}^{2} b \: {sin}^{2} a + {sin}^{2} a \: {sin}^{2} b \\ \\ = {sin}^{2} a \times ( {cos}^{2} b + {sin}^{2} b) \\ \\ = {sin}^{2} a$

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solve by PRINCE PATEL..

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