Question

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

Answer 1

Given

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

Divide numerator and denominator by $\cos \theta$

$=\frac{\frac{\sin \theta }{\cos \theta }-\frac{\cos \theta }{\cos \theta }+\frac{1}{\cos \theta }}{\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\cos \theta }-\frac{1}{\cos \theta }}$

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

Put $1={\sec }^2\theta -{\tan }^2\theta$ in numerator

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

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

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

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

$=\frac{(\sec \theta +\tan \theta )[1-\sec \theta +\tan \theta ]}{\tan \theta -\sec \theta -1}$

$=\frac{(\sec \theta +\tan \theta )[\tan \theta -\sec \theta +1]}{\tan \theta -\sec \theta -1}$

$=\sec \theta +\tan \theta$

Multiply and divide with $(\sec \theta -\tan \theta )$

$=(\sec \theta +\tan \theta )\times \frac{(\sec \theta -\tan \theta )}{(\sec \theta -\tan \theta )}$

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

$\because {\sec }^2\theta -{\tan }^2\theta =1$

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

Hence proved