Question

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

Answer 1

Prove :-

$\frac{1}{sec\theta -tan\theta }-\frac{1}{cos\theta }=\frac{1}{cos\theta }-\frac{1}{sec\theta -tan\theta }$

LHS

$\frac{1}{sec\theta -tan\theta }-\frac{1}{cos\theta }$

$\Rightarrow \frac{sec\theta +tan\theta }{se{c}^2\theta -ta{n}^2\theta }-\frac{1}{cos\theta }$

$\Rightarrow \frac{1}{cos\theta }+\frac{sin\theta }{cos\theta }-\frac{1}{cos\theta }$

$\Rightarrow tan\theta$

RHS :-

$\frac{1}{cos\theta }$ -$\frac{(sec\theta -tan\theta )}{se{c}^2\theta -ta{n}^2\theta }$

$\frac{1}{cos\theta }-\frac{1}{cos\theta }+tan\theta$

$tan\theta$

$\therefore$ LHS =RHS

$\therefore$ $\frac{1}{sec\theta -tan\theta }$ $-\frac{1}{cos\theta }=\frac{1}{cos\theta }$ -$\frac{1}{sec\theta -tan\theta }$