cos (A+B) ⋅ cos (A-B) is given by:

Question

cos (A+B) ⋅ cos (A-B) is given by: (A)  cos 2 A - cos 2 B (B)  cos (A2-B2) (C)  cos 2 A - sin 2 B (D)  sin 2 A- cos 2 B

Tardigrade Admin · Accepted Answer

cos (A+B) ⋅ cos (A-B)=( cos A cos B- sin A sin B) ( cos A cos B+ sin A sin B) = cos 2 A cos 2 B- sin 2 A sin 2 B = cos 2 A(1- cos A cos B+ sin A sin B) = cos 2 A- cos 2 B sin 2 B- sin 2 A sin 2 B = cos 2 A- sin 2 B( cos 2 A+ sin 2 A)= cos 2 A- sin 2 B

Q. $cos (A + B) \cdot cos (A - B)$ is given by:

$cos^{2} A - cos^{2} B$

$cos (A^{2} - B^{2})$

$cos^{2} A - sin^{2} B$

$sin^{2} A - cos^{2} B$

Q. cos(A+B)⋅cos(A−B) is given by:

cos2A−cos2B

cos(A2−B2)

cos2A−sin2B

sin2A−cos2B

cos(A+B)⋅cos(A−B)=(cosAcosB−sinAsinB) (cosAcosB+sinAsinB) =cos2Acos2B−sin2Asin2B =cos2A(1−cosAcosB+sinAsinB) =cos2A−cos2Bsin2B−sin2Asin2B =cos2A−sin2B(cos2A+sin2A)=cos2A−sin2B

Q. $cos (A + B) \cdot cos (A - B)$ is given by:

$cos^{2} A - cos^{2} B$

$cos (A^{2} - B^{2})$

$cos^{2} A - sin^{2} B$

$sin^{2} A - cos^{2} B$