Prove that:

Solution:

Here we have to prove that,

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

Firs we take the left hand side of the given equation

$L H S=\left(\frac{\sec \theta+\tan \theta-1}{\tan \theta-\sec \theta+1}\right)$

Now we are using the trigonometric identity

$1+\tan ^{2} \theta=\sec ^{2} \theta$

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

Therefore,

$L H S=\frac{\sec \theta+\tan \theta-\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{\tan \theta-\sec \theta+1}$

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

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

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

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

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

$=R H S$

Hence, $\left(\frac{\sec \theta+\tan \theta-1}{\tan \theta-\sec \theta+1}\right)=\frac{\cos \theta}{1-\sin \theta}$.