Question

Prove that: ${\displaystyle \frac{1}{\text{sec θ − tan θ}}=\text{sec θ + tan θ}}$

Answer 1

LHS = ${\displaystyle \frac{1}{\text{sec θ − tan θ}}}$

LHS = ${\displaystyle \frac{1}{\text{sec θ − tan θ}}\times \frac{\text{sec θ + tan θ}}{\text{sec θ + tan θ}}}$

LHS = ${\displaystyle \frac{\text{sec θ + tan θ}}{(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )}}$

LHS = ${\displaystyle \frac{\sec \theta +\tan \theta }{{\sec }^2\theta -{\tan }^2\theta } ...[(a+b)(a-b)=a^2-b^2]}$

LHS = ${\displaystyle \frac{\sec \theta +\tan \theta }{1} ...\{\begin{array}{l}1+{\tan }^2\theta ={\sec }^2\theta \\ \therefore {\sec }^2\theta -{\tan }^2\theta =1\end{array}}$

LHS = sec θ + tan θ

RHS = sec θ + tan θ

LHS = RHS

Hence proved.