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

`(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)`

A

`tan(A-B)`

B

`tan(A+B)`

C

`cot(A-B)`

D

`cot(A+B)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(sin^(2)A-sin^(2)B)/(sinAcosA-sinBcosB)=(2sin(A+B)sin(A-B))/(sin2A-sin2B)`
`=(2sin(A+B)sin(A-B))/(2sin(A-B)cos(A+B))`
`=tan(A+B)`
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