Question

Prove that :

$\frac{1}{ \sec( \alpha ) - \tan( \alpha ) } = \sec( \alpha ) + \tan( \alpha )$

Answer 1

We know, that sec^2 theta -tan^2 theta=1
From the Identity of a^2-b^2=(a+b)(a-b)
(sec theta + tan theta) (sec theta -tan theta)=1
sec theta +tan theta =1/sec theta - tan thets

Answer 2

Hey!

Here is your answer!

$to \: prove \\ \\ \frac{1}{sec \alpha - tan \alpha } = sec \alpha + tan \alpha \\ \\ proof \\ \\ lhs \\ \\ = \frac{1}{sec \alpha - tan \alpha } \\ \\ = \frac{1}{ \frac{1}{cos \alpha } - \frac{sin \alpha }{ cos \alpha } } \\ \\ = \frac{1}{ \frac{1 - sin \alpha }{cos \alpha } } \\ \\ = \frac{cos \alpha }{1 - sin \alpha } \\ \\ rationalising \: the \: denominator \\ \\ = \frac{cos \alpha (1 + sin \alpha )}{(1 - sin \alpha )(1 + sin \alpha )} \\ \\ = \frac{cos \alpha (1 + sin \alpha )}{1 - {sin}^{2} \alpha } \\ \\ using \: 1 - {sin}^{2} \alpha = {cos}^{2} \alpha \\ \\ = \frac{cos \alpha (1 + sin \alpha )}{ {cos}^{2} \alpha } \\ \\ = \frac{1 + sin \alpha }{cos \alpha } \\ \\ = \frac{1}{cos \alpha } + \frac{sin \alpha }{cos \alpha } \\ \\ = sec \alpha + tan \alpha \\ \\ = rhs \\ \\ hence \: proved$

Hope it's satisfactory!