Verify that \(\left(\frac{cos^3 \ θ - sin^3 \ θ}{cos \ θ - sin \ θ}\right) - \left(\frac{cos^3 \ θ + sin^3 \ θ}{cos \ θ + sin \ θ}\right) = 2 cos θ sin θ\)
Proof:
LHS \(= \frac{cos^3 \ θ - sin^3 \ θ}{cos \ θ - sin \ θ} - \frac{cos^3 \ θ + sin^3 \ θ}{cos \ θ + sin \ θ}\)
By applying the formula \(a^3+b^3=(\)\()(a^2\)\(+b^2)\) ,
\(a^3-b^3=(\)\()(a^2\)\(+b^2)\) then simplyfing we get,
\(= \left(cos^2 \ θ + cos \ θ \ sin \ θ + sin^2 \ θ\right) - \left(cos^2 \ θ - cos \ θ \ sin \ θ + sin^2 \ θ\right)\)
By the known trigonometric identity, we get:
\(= 2 \ cos \ θ \ sin \ θ\)
Hence, proved.