Question

Prove that $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }$, using the identity ${\sec }^2\theta =1+{\tan }^2\theta$.

Answer 1

Given : $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}$

Now divide both numerator and denominator with $\cos \theta$

We get,

$\Rightarrow \frac{\tan \theta -1+\sec \theta }{\tan \theta +1-\sec \theta }$

Now from the identity, put $1={\sec }^2\theta -{\tan }^2\theta$

$\Rightarrow \frac{\tan \theta +\sec \theta +{\tan }^2\theta -{\sec }^2\theta }{\tan \theta +1-\sec \theta }=\frac{(\tan \theta +sec\theta )(1+\tan \theta -\sec \theta )}{1+\tan \theta -\sec \theta }$

$=\tan \theta +\sec \theta =\frac{1}{sec\theta -\tan \theta }(\because (\sec \theta +\tan \theta \left)\right(\sec \theta -\tan \theta )=1\text{from the identity})$

Hence proved.