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Prove that ...

Question

Prove that $\frac{1 + csc A + cot A}{1 + csc A - cot A} = \frac{csc A + cot A - 1}{cot A - csc A + 1}$

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Solution

LHS:
$\frac{1 + csc A + cot A}{1 + csc A - cot A}$
$= \frac{sin A + c o s A + 1}{sin A - cos A + 1}$
$= (\frac{sin A + c o s A + 1}{sin A - cos A + 1})$ $(\frac{sin A + c o s A + 1}{sin A + cos A + 1})$
$= \frac{2 [1 + sin A + cos A + sin A cos A]}{2 {sin}^{2} A + 2 sin A} = \frac{1 + cos A}{sin A}$
RHS:
$\frac{- 1 + csc A + cot A}{1 - csc A - cot A}$
$= \frac{- sin A + c o s A + 1}{sin A + cos A - 1}$
$= (\frac{- sin A + c o s A + 1}{sin A + cos A - 1})$ $(\frac{- sin A + c o s A + 1}{- sin A + cos A + 1})$
$= \frac{2 [1 - sin A + cos A - sin A cos A]}{- 2 {sin}^{2} A + 2 sin A} = \frac{1 + cos A}{sin A}$
LHS $=$ RHS.

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Q. Prove that :

\frac{1}{csc A - cot A} - \frac{1}{sin A} = \frac{1}{sin A} - \frac{1}{csc A + cot A}

\frac{cot A + csc A - 1}{cot A - csc A + 1} = \frac{1 + cos A}{1 + sin A}

\frac{cos A - 1}{sin A} = \frac{cot A - 1 - csc A}{csc A + 1 + cot A}

\frac{cot A + csc A - 1}{cot A - csc A + 1} = \frac{1 + cos A}{sin A}

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