Question

$\frac{{\cot }^3\mathrm{\theta }·{\sin }^3\mathrm{\theta }}{{\left(\cos \mathrm{\theta }+\sin \mathrm{\theta }\right)}^2}+\frac{{\tan }^3\mathrm{\theta }·{\cos }^3\mathrm{\theta }}{{\left(\cos \mathrm{\theta }+\sin \mathrm{\theta }\right)}^2}=\frac{\sec \mathrm{\theta }\mathrm{cosec}\mathrm{\theta }-1}{\mathrm{cosec}\mathrm{\theta }+\sec \mathrm{\theta }}$

Answer 1

Dear student
$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{LHS}=\frac{{\cot }^3{\mathrm{\theta sin}}^3\mathrm{\theta }}{{\left(\mathrm{cos\theta }+\mathrm{sin\theta }\right)}^2}+\frac{{\tan }^3{\mathrm{\theta cos}}^3\mathrm{\theta }}{{\left(\mathrm{cos\theta }+\mathrm{sin\theta }\right)}^3} \\ =\frac{{\cos }^3\mathrm{\theta }+{\sin }^3\mathrm{\theta }}{{\left(\mathrm{cos\theta }+\mathrm{sin\theta }\right)}^2} \\ =\frac{\left(\mathrm{cos\theta }+\mathrm{sin\theta }\right)\left({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }-\mathrm{cos\theta }\mathrm{sin\theta }\right)}{{\left(\mathrm{cos\theta }+\mathrm{sin\theta }\right)}^2} \\ =\frac{1-\mathrm{cos\theta }\mathrm{sin\theta }}{\mathrm{cos\theta }+\mathrm{sin\theta }} \\ \mathrm{RHS}=\frac{\mathrm{sec\theta }\mathrm{cosec\theta }-1}{\mathrm{cosec\theta }+\mathrm{sec\theta }} \\ =\frac{{\displaystyle \frac{1}{\mathrm{cos\theta sin\theta }}}-1}{{\displaystyle \frac{1}{\mathrm{sin\theta }}}+{\displaystyle \frac{1}{\mathrm{cos\theta }}}} \\ =\frac{{\displaystyle \frac{1-\mathrm{sin\theta }\mathrm{cos\theta }}{\mathrm{cos\theta }\mathrm{sin\theta }}}}{{\displaystyle \frac{\mathrm{cos\theta }+\mathrm{sin\theta }}{\mathrm{cos\theta }\mathrm{sin\theta }}}} \\ =\frac{1-\mathrm{cos\theta }\mathrm{sin\theta }}{\mathrm{cos\theta }+\mathrm{sin\theta }} \\ \mathrm{Therefore},\mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved} \end{gathered}$$
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