Question

Prove that:

\[\sec\theta + \tan\theta = \frac{\cos\theta}{1 - \sin\theta}\]

Answer 1

L.H.S. = \[\sec\theta + \tan\theta\]

\[ = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\]

\[ = \frac{1 + \sin\theta}{\cos\theta}\]

= \[\frac{1+\sin\theta}{\cos\theta}\times\frac{1-\sin\theta}{1-\sin\theta}\]

= \[\frac{1^{2}-\sin^{2}\theta}{\cos\theta\bigl(1-\sin\theta\bigr)}\]

= \[\frac{1-\sin^{2}\theta}{\cos\theta\left(1-\sin\theta\right)}\]

= \[\frac{\cos^{2}\theta}{\cos\theta(1-\sin\theta)}\]           ...\[\begin{bmatrix}\because\sin^{2}\theta+\cos^{2}\theta=1\\\therefore1-\sin^{2}\theta=\cos^{2}\theta\end{bmatrix}\]

= \[\frac{\cos\theta}{1 - \sin\theta}\]

= R.H.S.

∴ \[\sec\theta + \tan\theta = \frac{\cos\theta}{1 - \sin\theta}\]