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Prove that:$\frac{cos\ A\ −\ sin\ A\ +\ 1}{cos\ A\ +\ sin\ A\ −\ 1} \ =\ cosec\ A\ +\ cot\ A$
Given: $\frac{cosA\ −\ sinA\ +\ 1}{cosA\ +\ sinA\ −\ 1} \ =\ cosecA\ +\ cotA$
To do: Here we have to prove that $\frac{cosA\ −\ sinA\ +\ 1}{cosA\ +\ sinA\ −\ 1} \ =\ cosecA\ +\ cotA$.
Solution:
Now,
$\frac{cosA\ −\ sinA\ +\ 1}{cosA\ +\ sinA\ −\ 1} \ =\ cosecA\ +\ cotA$
Dividing numerator and denominator of LHS by $sin\ A$:
$=\ \frac{\frac{cos\ A}{sin\ A} \ −\ \frac{sin\ A}{sin\ A} \ +\ \frac{1}{sin\ A}}{\frac{cos\ A}{sin\ A} \ +\ \frac{sin\ A}{sin\ A} \ −\ \frac{1}{sin\ A}}$
$=\ \frac{cot\ A\ −\ 1\ +\ cosec\ A}{cot\ A\ +\ 1\ -\ cosec\ A}$
$=\ \frac{cot\ A\ −\ 1\ +\ cosec\ A}{cot\ A\ +\ 1\ -\ cosec\ A}$
$=\ \frac{cot\ A\ +\ cosec\ A\ −\ 1}{cot\ A\ +\ 1\ -\ cosec\ A}$
$=\ \frac{cot\ A\ +\ cosec\ A\ −\ \left( cosec^{2} \ A\ -\ cot^{2} \ A\right)}{cot\ A\ +\ 1\ -\ cosec\ A}$
$=\ \frac{cot\ A\ +\ cosec\ A\ −\ \{( cosec\ A\ -\ cot\ A)( cosec\ A\ +\ cot\ A)\}}{cot\ A\ +\ 1\ -\ cosec\ A}$
$=\ \frac{( cot\ A\ +\ cosec\ A)\{1\ -\ ( cosec\ A\ -\ cot\ A)\}}{cot\ A\ +\ 1\ -\ cosec\ A}$
$=\ \frac{( cot\ A\ +\ cosec\ A)\{1\ -\ cosec\ A\ +\ cot\ A\}}{\{1\ +\ cot\ A\ -\ cosec\ A\}}$
$=\ \mathbf{cot\ A\ +\ cosec\ A}$
So, LHS is equal to RHS.
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