Answer:
Step-by-step explanation:
2 cos(A+45)cos (B-45)
=2[cos A cos 45-sin A sin 45][cos B cos 45+sin B sin 45]
=2[cos A ×1/√2-sin A ×1/√2][cos B×1/√2+sin B×1/√2]
=2[1/√2(cos A-sin A)][1/√2(cos B-sin B]
=2×1/√2×1/√2 (cos A-sin A)(cos B+sin B)
=cosA cos B+cos A sin B-sin A cos B-sin A sin B
=cos A cos B-sin A sin B-(sin A cos B-cos A sin B)
=cos (A+B)-sin (A-B)
i think there should be A in place of B.
then
=cos (A+A)-sin (A-A)
=cos 2A-sin 0
=cos 2A
as sin 0=0
