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