Question

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

Answer 1

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Answer 2

Proof  the problem :

L.H.S. = $\dfrac{cos\theta*cot\theta}{1+sin\theta}$

= $\dfrac{cos\theta * \dfrac{cos\theta}{sin\theta}*(1-sin\theta)}{(1+sin\theta)*(1-sin\theta)}$

[ by multiplying both the numerator and denominator by (1 - sinθ) ]

= $\dfrac{\dfrac{cos^{2}\theta}{sin\theta}*(1-sin\theta)}{1-sin^{2}\theta}$

= $\dfrac{\dfrac{cos^{2}\theta}{sin\theta}*(1-sin\theta)}{cos^{2}\theta}$

[ since $sin^{2}\theta+cos^{2}\theta=1$ ]

= $\dfrac{1-sin\theta}{sin\theta}$

[ by cancelling $cos^{2}\theta$ from both the numerator and the denominator ]

= $\dfrac{1}{sin\theta} - \dfrac{sin\theta}{sin\theta}$

(using the formula : $\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c})$

= cosecθ - 1

= R.H.S.

Hence, proved.