Question

Cos^2b-sin^2a/sina.cosa+sinb.cosb=cot(a+b)

Answer 1

To find :

To prove ( cos^2b - sin^2a ) / ( sina.cosa + sinb.cosb ) = cot (a+b)

Solution :

L.H.S. :

( cos^2b - sin^2a ) / ( sina.cosa + sinb.cosb )

=> 2 ( cos^2b - sin^2a ) / 2 ( sina.cosa + sinb.cosb )

=> ( 2 cos^2b - 2 sin^2a ) / ( sin 2a + sin 2b )

=> ( 2 cos^2b - 2 sin^2a + 1 - 1 ) / 2 sin (a+b) cos (a-b)

=> ( 2 cos^2b -1 + 1 - 2 sin^2a ) / 2 sin (a+b) cos (a-b)

=> ( cos 2b + cos 2a ) / 2 sin (a+b) cos (a-b)

=> ( 2 cos (a+b) cos (a-b) ) / 2 sin (a+b) cos (a-b)

=> cos (a+b) / sin (a+b) = cot (a+b) = R.H.S.

Hence proved , ( cos^2b - sin^2a ) / ( sina.cosa + sinb.cosb ) = cot (a+b)