Prove that: $(sin²A-sin²B)/((sin A cos A)-(sin B cos B))=tan(A+B)$ ?

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Answer 1 · 2018-06-07T22:48:56

$LHS=(sin²A-sin²B)/((sin A cos A)-(sin B cos B))$

$=(2sin(A+B)sin(A-B))/((2sin A cos A)-(2sin B cos B))$

$=(2sin(A+B)sin(A-B))/(sin 2A -sin2 B)$

$=(2sin(A+B)sin(A-B))/(2cos(A+B) sin(A- B))$

$=tan(A+B)=RHS$

Prove that: (sin²A-sin²B)/((sin A cos A)-(sin B cos B))=tan(A+B)(sin²A-sin²B)/((sin A cos A)-(sin B cos B))=tan(A+B)?