Question

sin theta minus cos theta + 1 / sin theta + cos-theta -1 = 1/ sec theta -tan theta
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Answer 1

Solution:

$\sf{\implies \dfrac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}= \dfrac{1}{\sec \theta - \tan \theta}}$

We take LHS part,

$\sf{\implies \dfrac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}}$

Now we will do Rationalize,

$\sf{\implies \dfrac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}\times \dfrac{\sin \theta + \cos \theta + 1}{\sin \theta + \cos \theta + 1}}$

$\sf{\implies \dfrac{\sin^{2}\theta + \sin \theta \cos \theta + \sin \theta - \sin \theta \cos \theta -\cos^{2} \theta - \cos \theta + \sin \theta +\cos\theta +1}{(\sin \theta +\cos \theta)^{2}-1}}$

$\sf{\implies \dfrac{2\sin^{2}\theta + 2\sin \theta}{(\sin^{2}\theta + \cos^{2}\theta)+2\sin\theta \cos \theta -1}}$

$\sf{\implies \dfrac{2\sin^{2}\theta + 2\sin \theta}{1+2\sin \theta \cos \theta-1}}$

$\sf{\implies \dfrac{2\sin \theta(1+\sin\theta)}{2\sin\theta \cos\theta}}$

$\sf{\implies \dfrac{1+\sin\theta}{\cos\theta}}$

$\sf{\implies \sec\theta+\tan \theta}$

$\sf{\implies \sec \theta +\tan \theta \times \dfrac{\sec \theta-\tan \theta}{\sec \theta -\tan \theta}}$

$\sf{\implies \dfrac{(\sec^{2}\theta-\tan^{2}\theta)}{\sec \theta -\tan\theta}}$

$\sf{\implies \dfrac{1}{\sec\theta-\tan\theta}}$

Hence Proved!!!

Answer 2

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$\boxed{Explained\:Answer}$

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