In any triangle `A B C ,sin^2A-sin^2B+sin^2C` is always equal to (A) `2sinAsinBcosC` (B) `2sinAcosBsinC` (C) `2sinAcosBcosC` (D) `2sinAsinBsinC`

Explanation

`sin^2A-sin^2B+sin^2C`

As `sin^2A-sin^2B = sin(A+B)sin(A-B) `, it becomes,

`=sin(A+B)sin(A-B) +sin^2C`

`=sin(pi-C)sin(A-B) +sin^2C`

`=sinCsin(A-B) +sin^2C`

`=sinC(sin(A-B)+sinC)`

`=sinC(sin(A-B)+sin(pi-(A+B))`

`=sinC(sin(A-B)+sin(A+B))`

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