Question

Prove: (cosec​Θ - sin​Θ)(sec​Θ - cos​Θ) = 1/tan​Θ + cot​Θ

Answer 1

$$\begin{gathered} \mathrm{LHS}=\left(\mathrm{cosec}\mathrm{\theta }-\sin \mathrm{\theta }\right)\left(\sec \mathrm{\theta }-\cos \mathrm{\theta }\right) \\ =\left[\frac{1}{\sin \mathrm{\theta }}-\sin \mathrm{\theta }\right]\left[\frac{1}{\cos \mathrm{\theta }}-\cos \mathrm{\theta }\right] \\ =\left[\frac{1-{\sin }^2\mathrm{\theta }}{\sin \mathrm{\theta }}\right]\left[\frac{1-{\cos }^2\mathrm{\theta }}{\cos \mathrm{\theta }}\right] \\ =\frac{{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta }}\times \frac{{\sin }^2\mathrm{\theta }}{\cos \mathrm{\theta }} \\ =\sin \mathrm{\theta }.\cos \mathrm{\theta } \\ \mathrm{RHS}=\frac{1}{\tan \mathrm{\theta }+\cot \mathrm{\theta }} \\ =\frac{1}{{\displaystyle \frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}}+{\displaystyle \frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}}} \\ =\frac{\sin \mathrm{\theta }.\cos \mathrm{\theta }}{{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }} \\ =\sin \mathrm{\theta }.\cos \mathrm{\theta } \\ \mathrm{So},\mathrm{LHS}=\mathrm{RHS} \end{gathered}$$