Question

Plz ans this ques urgently..... This is vry imp..... Plz Plz Solve it

Answer 1

✍HERE IS YOUR SOLUTION...⤵⤵
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$\underline{SOLUTION}$ ➡

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

$Now,$

$L.H.S.= \frac{ \sin \theta - \cos \: \theta + 1 }{ \sin \: \theta + \cos \theta - 1} \\ \\ = \frac{ \tan\theta - 1 + sec \: \theta}{tan \: \theta + 1 - \sec\theta} \\ \\ = \frac{(tan \: \theta + sec \:\theta) - 1 }{(tan \: \theta - sec \:\theta) + 1 } \\ \\ = \frac{[(tan \: \theta + sec \:\theta) - 1](tan \: \theta \: - sec \: \theta)}{[(tan \: \theta - sec \:\theta) + 1 ](tan \: \theta \: - sec \: \theta)} \\ \\ = \frac{( \tan {}^{2} \theta - \sec{}^{2}\theta) - ( \tan \theta - sec \: \theta) }{[(tan \: \theta - sec \:\theta) + 1 ](tan \: \theta \: - sec \: \theta)} \\ \\ = \frac{(tan {}^{2} \: \theta - 1 - tan {}^{2}\theta) - ( \tan \theta - sec \: \theta) }{[(tan \: \theta - sec \:\theta) + 1 ](tan \: \theta \: - sec \: \theta)} \\ \\ = \frac{ - 1 - tan \:\theta + sec \: \theta }{(tan \: \theta - sec \:\theta + 1 )(tan \: \theta \: - sec \: \theta)} \\ \\ = \frac{ - 1(1 + tan \: \theta \: - sec \: \theta)}{(1 + tan \: \theta - sec \:\theta)(tan \: \theta \: - sec \: \theta)} \\ \\ = \frac{ - 1}{tan \: \theta - sec \: \theta} \\ \\ = \frac{1}{sec \: \theta - tan \: \theta} \\ \\ = R.H.S.$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Hence \: \: Proved \: .$

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