Q4(i):

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

Given: An identity involving trigonometric functions of an acute angle $\theta$, specifically $(\text{cosec } \theta - \cot \theta)^2$.

To Prove: $(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$

Step 1: Expressing the Left Hand Side (LHS) in terms of Sine and Cosine.

The given expression is $LHS = (\text{cosec } \theta - \cot \theta)^2$.

We use the fundamental trigonometric identities:
$\text{cosec } \theta = \frac{1}{\sin \theta}$
$\cot \theta = \frac{\cos \theta}{\sin \theta}$

Substituting these into the LHS:

$LHS = \left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right)^2$

Step 2: Simplifying the fraction inside the square.

Since the denominators are the same, we combine the terms:

$LHS = \left( \frac{1 - \cos \theta}{\sin \theta} \right)^2$

$LHS = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$

Step 3: Applying the Pythagorean Identity.

We know the identity $\sin^2 \theta + \cos^2 \theta = 1$, which implies $\sin^2 \theta = 1 - \cos^2 \theta$.

Substituting this into the denominator:

$LHS = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$

Step 4: Factoring the denominator.

The denominator is in the form of a difference of squares, $a^2 - b^2 = (a - b)(a + b)$, where $a = 1$ and $b = \cos \theta$.

$1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$

Substituting this back into the expression:

$LHS = \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}$

Step 5: Final Cancellation.

Canceling the common term $(1 - \cos \theta)$ from the numerator and the denominator:

$LHS = \frac{1 - \cos \theta}{1 + \cos \theta}$

Since $LHS = RHS$, the identity is proven.

Final Answer: Since the Left Hand Side simplifies to $\frac{1 - \cos \theta}{1 + \cos \theta}$, which is equal to the Right Hand Side, the identity $(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$ is proven.