Question

Prove that: $(\mathrm{sin\theta }-\mathrm{cosec\theta })(\mathrm{cos\theta }-\mathrm{sec\theta })=\frac{1}{(\mathrm{tan\theta }+\mathrm{cot\theta })}.$

Answer 1

$\mathrm{LHS}=\left(\mathrm{sin\theta }-\mathrm{cosec\theta }\right)\left(\mathrm{cos\theta }-\mathrm{sec\theta }\right)$
$$\begin{gathered} =\left(\mathrm{sin\theta }-\frac{1}{\mathrm{sin\theta }}\right)\left(\mathrm{cos\theta }-\frac{1}{\mathrm{cos\theta }}\right) \\ =\frac{\left({\sin }^2\mathrm{\theta }-1\right)}{\mathrm{sin\theta }}\times \frac{\left({\cos }^2\mathrm{\theta }-1\right)}{\mathrm{cos\theta }} \end{gathered}$$
$=\frac{\left\{-\left(1-{\sin }^2\mathrm{\theta }\right)\right\}\times \left\{-\left(1-{\cos }^2\mathrm{\theta }\right)\right\}}{\mathrm{sin\theta cos\theta }}$
Now, (1- sin²θ) = cos²θ and (1- cos²θ) = sin²θ
$$\begin{gathered} \text{Therefore, we have:}\frac{\left(-{\cos }^2\mathrm{\theta }\right)\left(-{\sin }^2\mathrm{\theta }\right)}{\mathrm{sin\theta cos\theta }} \\ =\frac{{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }}{\mathrm{sin\theta cos\theta }} \\ =\mathrm{cos\theta sin\theta } \end{gathered}$$
$\mathrm{RHS}=\frac{1}{\mathrm{tan\theta }+\mathrm{cot\theta }}$
$=\frac{1}{{\displaystyle \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}+{\displaystyle \frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}}$
$=\frac{\mathrm{cos\theta sin\theta }}{{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }}\left[\mathrm{Since}{\sin }^2\mathrm{\theta }+{\cos }^2=1\right]$
$=c\mathrm{os\theta sin\theta }$
Hence, LHS = RHS