Q4(v):

Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (v) $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \text{cosec } A + \cot A$, using the identity $\text{cosec}^2 A = 1 + \cot^2 A$.

Given: The trigonometric expression $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$.

To Prove: $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A$, using the identity $\text{cosec}^2 A = 1 + \cot^2 A$.

Step 1: Manipulating the Left-Hand Side (LHS)

We begin with the expression:
$LHS = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$

To express the terms in $\text{cosec } A$ and $\cot A$, we divide both the numerator and the denominator by $\sin A$.
$LHS = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}}$

Step 2: Simplifying using Trigonometric Ratios

Recall the definitions: $\cot A = \frac{\cos A}{\sin A}$ and $\text{cosec } A = \frac{1}{\sin A}$.
Substituting these into the expression:
$LHS = \frac{\cot A - 1 + \text{cosec } A}{\cot A + 1 - \text{cosec } A}$

Step 3: Applying the Trigonometric Identity

We are given the identity $\text{cosec}^2 A = 1 + \cot^2 A$. This can be rearranged as:
$1 = \text{cosec}^2 A - \cot^2 A$
We substitute this value of $1$ into the denominator of our expression:
$LHS = \frac{\cot A + \text{cosec } A - 1}{\cot A - \text{cosec } A + (\text{cosec}^2 A - \cot^2 A)}$

Step 4: Factoring the Denominator

Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$, we factor $(\text{cosec}^2 A - \cot^2 A)$:
$LHS = \frac{\cot A + \text{cosec } A - 1}{\cot A - \text{cosec } A + (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}$

To simplify, we factor out $(\cot A + \text{cosec } A)$ from the denominator. Note that $(\cot A - \text{cosec } A) = -(\text{cosec } A - \cot A)$:
$LHS = \frac{\cot A + \text{cosec } A - 1}{-(\text{cosec } A - \cot A) + (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}$

Step 5: Final Cancellation

Factor out $(\text{cosec } A - \cot A)$ in the denominator:
$LHS = \frac{\cot A + \text{cosec } A - 1}{(\text{cosec } A - \cot A) [(\text{cosec } A + \cot A) - 1]}$

Since the term $(\cot A + \text{cosec } A - 1)$ appears in both the numerator and the denominator, they cancel out:
$LHS = \frac{1}{\text{cosec } A - \cot A}$

To reach the final form, multiply the numerator and denominator by the conjugate $(\text{cosec } A + \cot A)$:
$LHS = \frac{1 \cdot (\text{cosec } A + \cot A)}{(\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}$
$LHS = \frac{\text{cosec } A + \cot A}{\text{cosec}^2 A - \cot^2 A}$

Since $\text{cosec}^2 A - \cot^2 A = 1$:
$LHS = \text{cosec } A + \cot A = RHS$

Final Answer: Since the Left-Hand Side equals the Right-Hand Side, the identity $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A$ is proved.