Question

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

Answer 1

Solving the function using trigonometric identities:

As we have $\frac{(\sin \theta -\cos \theta +1)}{(\sin \theta +\cos \theta -1)}=\frac{1}{(sec\theta -\tan \theta ).}$

LHS $=$$\frac{(\sin \theta –\cos \theta +1)}{(\sin \theta +\cos \theta –1)}$

Dividing the numerator and denominator by $\cos \theta$

$\frac{\left({\displaystyle \frac{\sin \theta }{\cos \theta }}\right)–\left({\displaystyle \frac{\cos \theta }{\cos \theta }}\right)+\left({\displaystyle \frac{1}{\cos \theta }}\right)}{\left(\frac{\sin \theta }{\cos \theta }\right)+\left(\frac{\cos \theta }{\cos \theta }\right)–\left(\frac{1}{\cos \theta }\right)}$

$=\frac{(\tan \theta –1+sec\theta )}{(\tan \theta +1–sec\theta )}$

Multiplying and dividing by $(\tan \theta –sec\theta ),$

$$\begin{gathered} =\frac{(\tan \theta +sec\theta –1)}{(\tan \theta –sec\theta +1)}\times \frac{(\tan \theta –sec\theta )}{(\tan \theta –sec\theta )}\left[(\tan \theta +sec\theta )\left(\right(\tan \theta –sec\theta ={\tan }^2\theta –se{c}^2\theta )\right] \\ =\frac{\left[\right({\tan }^2\theta –se{c}^2\theta –(\tan \theta –sec\theta )]}{\left[\right(\tan \theta –sec\theta +1)(\tan \theta –sec\theta \left)\right]} \end{gathered}$$

Using the identity $se{c}^2\theta –{\tan }^2\theta =1,$

$$\begin{gathered} =\frac{(-1–\tan \theta +sec\theta )}{\left[\right(\tan \theta –sec\theta +1\left)\right(\tan \theta –sec\theta \left)\right]} \\ =\frac{-1}{(\tan \theta –sec\theta )} \\ =\frac{1}{(sec\theta –\tan \theta )} \end{gathered}$$

$=$ RHS

Hence proved.