Question

(a) show that \\(\frac{\sin \theta + 2 \cos \theta}{\cos \theta - 2 \sin \theta} - \frac{\sin \theta - 2 \cos \theta}{\cos \theta + 2 \sin \theta} \equiv \frac{4}{5 \cos^2 \theta - 4}\\).

Explanation:

Combine the fractions over a common denominator

Using the Trigonometric Identities knowledge point

$$
LATEXBLOCK0
$$

Expand the numerator terms

Using the Trigonometric Identities knowledge point

$$
LATEXBLOCK1
$$

Simplify the numerator

Using the Trigonometric Identities knowledge point

$$
LATEXBLOCK2
$$

Simplify the denominator

Using the Trigonometric Identities knowledge point

$$
LATEXBLOCK3
$$

Reconstruct the final expression

Using the Trigonometric Identities knowledge point

$$
LATEXBLOCK4
$$

Answer:

To show that:

$$ \frac{\sin \theta + 2 \cos \theta}{\cos \theta - 2 \sin \theta} - \frac{\sin \theta - 2 \cos \theta}{\cos \theta + 2 \sin \theta} \equiv \frac{4}{5 \cos^2 \theta - 4} $$

We combine the left-hand side (LHS) over a common denominator:

$$ \text{LHS} = \frac{(\sin\theta + 2\cos\theta)(\cos\theta + 2\sin\theta) - (\sin\theta - 2\cos\theta)(\cos\theta - 2\sin\theta)}{(\cos\theta - 2\sin\theta)(\cos\theta + 2\sin\theta)} $$

Expanding the numerator:

$$ (\sin\theta + 2\cos\theta)(\cos\theta + 2\sin\theta) = 2(\sin^2\theta + \cos^2\theta) + 5\sin\theta\cos\theta = 2 + 5\sin\theta\cos\theta $$

$$ (\sin\theta - 2\cos\theta)(\cos\theta - 2\sin\theta) = -2(\sin^2\theta + \cos^2\theta) + 5\sin\theta\cos\theta = -2 + 5\sin\theta\cos\theta $$

$$ \text{Numerator} = (2 + 5\sin\theta\cos\theta) - (-2 + 5\sin\theta\cos\theta) = 4 $$

Expanding and simplifying the denominator using \(\sin^2\theta = 1 - \cos^2\theta\):

$$ \text{Denominator} = \cos^2\theta - 4\sin^2\theta = \cos^2\theta - 4(1 - \cos^2\theta) = 5\cos^2\theta - 4 $$

Putting it all together:

$$ \text{LHS} = \frac{4}{5\cos^2\theta - 4} = \text{RHS} $$