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(sin^2A-sin^2B)/(sinAcosA-sinBcosB)is eq...

`(sin^2A-sin^2B)/(sinAcosA-sinBcosB)`is equal to

A

`"tan"(A-B)`

B

`"tan"(A+B)`

C

`cot(A-B)`

D

`cot(A+B)`

Text Solution

Verified by Experts

`(sin^2A-sin^2B)/(sinAcosA - sinBcosB)`
`=(2(sin^2A-sin^2B))/(2sinAcosA - 2sinBcosB)`
`=(2(sin(A+B)sin(A-B)))/(sin2A - sin2B)`
`=(2(sin(A+B)sin(A-B)))/(2sin(A-B)cos(A+B))`
`=sin(A+B)/cos(A+B)`
`=tan(A+B)`
So, option `(b)` is the correct option.
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