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.