### Proving the Identities
#### Identity (i)
Let's start by proving:
[tex]\[
\frac{1+\cos \theta}{1-\cos \theta} = (\operatorname{cosec} \theta + \cot \theta)^2
\][/tex]
First, express [tex]\(\operatorname{cosec} \theta\)[/tex] and [tex]\(\cot \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[
\operatorname{cosec} \theta = \frac{1}{\sin \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}
\][/tex]
Thus,
[tex]\[
\operatorname{cosec} \theta + \cot \theta = \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = \frac{1 + \cos \theta}{\sin \theta}
\][/tex]
Squaring both sides,
[tex]\[
(\operatorname{cosec} \theta + \cot \theta)^2 = \left( \frac{1 + \cos \theta}{\sin \theta} \right)^2 = \frac{(1 + \cos \theta)^2}{\sin^2 \theta}
\][/tex]
Therefore, we have:
[tex]\[
(\operatorname{cosec} \theta + \cot \theta)^2 = \frac{(1 + \cos \theta)^2}{1 - \cos^2 \theta} \quad \text{(since \(\sin^2 \theta = 1 - \cos^2 \theta\))}
\][/tex]
We simplify:
[tex]\[
\frac{(1 + \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 + \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 + \cos \theta}{1 - \cos \theta}
\][/tex]
This proves the given identity:
[tex]\[
\boxed{\frac{1+\cos \theta}{1-\cos \theta} = (\operatorname{cosec} \theta + \cot \theta)^2}
\][/tex]
#### Identity (ii)
Next, we need to prove:
[tex]\[
\cos x + \cos 2x + \cos 3x = \cos 2x(2 \cos x + 1)
\][/tex]
First, use the sum-to-product identities:
[tex]\[
\cos x + \cos 3x = 2 \cos \left( \frac{4x}{2} \right) \cos \left( \frac{-2x}{2} \right) = 2 \cos 2x \cos(-x) = 2 \cos 2x \cos x
\][/tex]
Add [tex]\(\cos 2x\)[/tex] to both sides:
[tex]\[
\cos x + \cos 2x + \cos 3x = 2 \cos 2x \cos x + \cos 2x
\][/tex]
Factor out [tex]\(\cos 2x\)[/tex]:
[tex]\[
2 \cos 2x \cos x + \cos 2x = \cos 2x (2 \cos x + 1)
\][/tex]
This proves the given identity:
[tex]\[
\boxed{\cos x + \cos 2x + \cos 3x = \cos 2x (2 \cos x + 1)}
\][/tex]
### Eliminating [tex]\(\theta\)[/tex]
Given the equations:
[tex]\[
x = \tan \theta, \quad y = 2 \sec^2 \theta - 1
\][/tex]
Recall the trigonometric identity for [tex]\(\sec \theta\)[/tex]:
[tex]\[
\sec^2 \theta = 1 + \tan^2 \theta
\][/tex]
Substitute [tex]\(\sec^2 \theta\)[/tex] with [tex]\(1 + \tan^2 \theta\)[/tex]:
[tex]\[
y = 2 (1 + \tan^2 \theta) - 1 = 2(1 + x^2) - 1 = 2 + 2x^2 - 1 = 2x^2 + 1
\][/tex]
Thus,
[tex]\[
y = 2x^2 + 1
\][/tex]
This expresses [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] and eliminates [tex]\(\theta\)[/tex]:
[tex]\[
\boxed{y = 2x^2 + 1}
\][/tex]
