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Q. Show that ,
sinAcosA + sinAcosB/ sin²A - sin²B = cot ( A - B )
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sinAcosA + sinBcosB/ sin²A - sin²B = cot( A - B )
L.H.S
=sinAcosA+sinBcosB/ sin²A-sin²B *2/2
= sin2A+sin2B/2 sin²A-sin²B
=2sin(2A+2B/2)cos(2A-2B/2) / 2sin(A+B)sin(A-B)
=2sin(A+B)cos(A-B)/ 2sin(A+B)sin(A-B)
=cos(A-B)/sin(A-B)
=cot(A-B)
= R.H.S
L.H.S
=sinAcosA+sinBcosB/ sin²A-sin²B *2/2
= sin2A+sin2B/2 sin²A-sin²B
=2sin(2A+2B/2)cos(2A-2B/2) / 2sin(A+B)sin(A-B)
=2sin(A+B)cos(A-B)/ 2sin(A+B)sin(A-B)
=cos(A-B)/sin(A-B)
=cot(A-B)
= R.H.S
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