If cos a + cos B = 0 = sin a sin B then
cos (a – B) =
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Answer:
Given that: cos A + cos B = 0
Hence, cos B = - cos A —(1)
Given that: sin A + sin B = 0
Hence, sin B = - sin A — (2)
Now,
L.H.S. = cos 2A + cos 2B
= (cos^2 A - sin^2 A) + (cos^2 B - sin^2 B)
= (cos^2 A - sin^2 A) + ((- cos A)^2 - (-sin A)^2) — by substituting based on (1) & (2)
= (cos^2 A - sin^2 A) + (cos^2 A - sin^2 A)
= 2 (cos^2 A - sin^2 A)
= 2 (cos A . cos A - sin A . sin A)
= 2 (cos A . (-cos B) - sin A . (-sin B)) — by substituting based on (1) & (2)
= 2 (- cos A cos B + sin A sin B)
= -2(cos A cos B - sin A sin B)
= -2 cos (A+B)
= 2 cos (π+A+B)
= R.H.S.
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Hope this will help you...
Thank you
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