Question

one upon sec theta minus tan theta equal to sec theta plus tan theta

Answer 1

Step-by-step explanation:

We have,

$\rm \dfrac{1}{\sec \theta - \tan \theta}$

$\rm \dfrac{1}{\sec \theta - \tan \theta}× \dfrac{\sec \theta + \tan \theta}{\sec \theta +\tan \theta}$

$\rm \dfrac{\sec \theta + \tan \theta}{\sec ^2 \theta -\tan ^2 \theta}$

$\rm \dfrac{\sec \theta + \tan \theta}{1}$

$\rm = \sec θ + \tan \theta$

Answer 2

$\bf\frac{1}{sec \: \theta - tan \: \theta} \\ \\ \bf \: = \: \frac{1}{sec \: \theta - tan \: \theta} \: \: \times \: \frac{sec \: \theta + tan \: \theta}{sec \: \theta + tan \: \theta} \\ \\ \bf \: = \: \frac{sec \: \theta + tan \: \theta}{sec^{2} \: \theta - tan^{2} \: \theta} \\ \\ \bf \: = \: \frac{sec \: \theta + tan \: \theta}{1} \\ \\ \bf \: = \: sec \: \theta + tan \: \theta$