Prove that \(\frac{(1 + cot \ A + tan \ A)(sin \ A - cos \ A)}{sec^3 \ A - cosec^3 \ A} = sin^2 \ A cos^2 \ A\)
Proof:
\(LHS = \frac{(1 + cot \ A + tan \ A)(sin \ A - cos \ A)}{sec^3 \ A - cosec^3 \ A}\)
We know that, \(cot \ A=\) ,
\(tan \ A=\)
\(sec A=\)
\(cosec A=\)
By applying the above ratios, the identity \(a^3-b^3=(\)\()a^2+ab+b^2\) and symplifying then we get,