Question

`(sin^2A-sin^2B)/(sinAcosA-sinBcosB)`is equal to (a)`"tan"(A-B)` (b) `"tan"(A+B)` `cot(A-B)` (d) `cot(A+B)`

Answer 1

`(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.