Question

prove that one upon cosec theta + cot theta minus one upon sin theta equals to one upon sin theta minus one by cos theta minus cos theta

Answer 1

hope it helps many. easy method

Answer 2

Answer:

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

Step-by-step explanation:

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

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

From equations (1) and (2) it follows that ,

LHS = RHS

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