Question

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

Answer 1

$\frac{1}{(\mathrm{sec\theta }-\mathrm{tan\theta })}-\frac{1}{\mathrm{cos\theta }}=\frac{1}{\mathrm{cos\theta }}-\frac{1}{(\mathrm{sec\theta }+\mathrm{tan\theta })}$

$$\begin{gathered} \text{LHS=}\frac{1}{\mathrm{sec\theta }-\mathrm{tan\theta }}-\frac{1}{\mathrm{cos\theta }} \\ \text{=}\frac{(\mathrm{sec\theta }+\mathrm{tan\theta })}{(\mathrm{sec\theta }-\mathrm{tan\theta })(\mathrm{sec\theta }+\mathrm{tan\theta })}-\mathrm{sec\theta }\left(\begin{array}{l}\text{Multipying the numerator and}\\ \text{denominator by}(\mathrm{sec\theta }+\mathrm{tan\theta })\end{array}\right) \\ \text{=}\frac{\mathrm{sec\theta }+\mathrm{tan\theta }}{{\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }}-\mathrm{sec\theta } \\ \text{=sec}\mathrm{\theta }\text{+tan}\mathrm{\theta }-\mathrm{sec\theta }[\because {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1] \\ \text{=}\mathrm{tan\theta } \\ \text{RHS=}\frac{1}{\mathrm{cos\theta }}-\frac{1}{\mathrm{sec\theta }+\mathrm{tan\theta }} \\ \text{=sec}\mathrm{\theta }-\frac{(\mathrm{sec\theta }-\mathrm{tan\theta })}{{\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }}\left(\begin{array}{l}\text{Multipying the numerator and}\\ \text{denomenator by}(\mathrm{sec\theta }-\mathrm{tan\theta })\end{array}\right) \\ \text{=}\mathrm{sec\theta }+\mathrm{tan\theta }-\mathrm{sec\theta }[\because {\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1] \\ \text{=}\mathrm{tan\theta } \\ \therefore \text{LHS=RHS} \\ \mathrm{Hence}\mathrm{Proved} \end{gathered}$$