Question

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

Answer 1

LHS = (cosec θ − sin θ)(sec θ − cos θ)
$$\begin{gathered} =\left(\frac{1}{\mathrm{sin\theta }}-\mathrm{sin\theta }\right)\left(\frac{1}{\mathrm{cos\theta }}-\cos \mathrm{\theta }\right) \\ =\frac{\left(1-{\sin }^2\mathrm{\theta }\right)}{\mathrm{sin\theta }}\times \frac{\left(1-{\cos }^2\mathrm{\theta }\right)}{\mathrm{cos\theta }} \end{gathered}$$
$=\frac{{\cos }^2{\mathrm{\theta sin}}^2\mathrm{\theta }}{\mathrm{sin\theta cos\theta }}$ [Since (1 - sin²θ) = cos²θ and (1 - cos²θ) = sin²θ]
$=\cos \mathrm{\theta sin\theta }$
$\mathrm{RHS}=\frac{1}{\mathrm{tan\theta }+\mathrm{cot\theta }}$
$$\begin{gathered} =\frac{1}{{\displaystyle \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}}+{\displaystyle \frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}}} \\ =\frac{\mathrm{cos\theta sin\theta }}{\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)} \end{gathered}$$
$=\frac{\mathrm{cos\theta sin\theta }}{1}=\mathrm{cos\theta sin\theta }$ [ Since sin²θ + cos²θ = 1]
Hence, LHS = RHS