Question

(1+cotФ+tanФ)(sinФ-cosФ)/sec∧3Ф-cosec∧3Ф=sin∧2Фcos∧2Ф

Answer 1

$\frac {(1+\frac{Cos\theta}{sin \theta}+\frac{Sin\theta}{Cos\theta})(sin\theta-cos\theta)}{\frac{1}{cos^3\theta} - \frac{1}{SIn^3\theta}} \\ \\ \\ \frac{(Sin\theta\ Cos\theta+Cos^2\theta+Sin^2\theta)(Sin\theta-Cos\theta)Sin^3\theta\ Cos^3\theta}{Sin\theta\ Cos\theta (Sin^3\theta - Cos^3\theta)} \\ \\ \frac{(Sin\theta\ Cos\theta+1)(Sin\theta-Cos\theta)Sin^2\theta\ Cos^2\theta}{(Sin\theta-Cos\theta)(Sin^2\theta+Sin\theta\ Cos\theta+Cos^2\theta)}$

Canceling terms and simplifying we get

$Sin^2\theta\ \ Cos^2\theta$