Question

If $\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{m}{\sec \theta -\tan \theta },$ Find $m$

Answer 1

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

multiply and divide by $\cos \theta$

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

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

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

multiply and divide by $\sec \theta -\tan \theta$

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

$=\frac{1}{\tan \theta -\sec \theta }$ [since$1={\sec }^2\theta -{\tan }^2\theta$]

$=m=1$