Question

cos theta-sin theta+1/cos theta+sin theta-1 =cosec theta+cot theta​

Answer 1

correct question :

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

Answer:

Take L.H.S

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

Dividing the Numerator and Denominator by,

$sin\theta$

we get ,

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

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Using Identity:

${coses }^{2} \theta = 1 + {cot}^{2} \theta \: \\ \\ {cosec}^{2} \theta - {cot}^{2} \theta = 1$

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$= > \frac{(cosec \theta + cot \theta) - ( {cosec}^{2}\theta - {cot}^{2}\theta}{1 + cot \theta - cosec \theta}$

$\boxed{(a + b)( a- b) = {a}^{2} - {b}^{2}}$

$= > \frac{(cose c\theta + cot \theta) - (cosec \theta + cot \theta)(cosec \theta - cot \theta) }{1 + cot \theta - cosec \theta} \\ \\ take \: commom ...(cosec \theta + cot \theta) \\ \\ = > \frac{(cosec \theta + cot \theta) \times1 -( cosec \theta - cot \theta) }{1 + cot \theta - cosec \theta} \\ \\ = > \frac{(cosec \theta + cot \theta) \times (1 + cot \theta - cosec \theta)}{1 + cot \theta - cosec \theta} \\ \\ = > cosec \theta + cot \theta = r.h.s$

Hence L.H.S = R.H.S