6. (i) Using the identity for tan(A ± B), solve, for −90° < x < 90°, tan 2x + tan 32° 1 − tan 2x tan 32° = 5 Give your answers, in degrees, to 2 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 5

We start with the equation:

$$\tan(30° − 45°) = \frac{\tan(30°) − 1}{1 + \tan(30°)}$$

First, we know that:\n\tan(30°) = \frac{1}{\sqrt{3}} ≈ 0.5774. Then substituting this into the equation gives us:

$$\tan(30° − 45°) = \frac{0.5774 − 1}{1 + 0.5774} = \frac{-0.4226}{1.5774} = -0.2686$$

Thus, we can confirm:

$$\tan(-15°) ≈ -0.2679 \text{ (approximately equal)}$$

From this, we state that θ ≠ (60n + 45)°, where n ∈ ℤ.

Starting from the equation:

$$(1 + \tan(30°))(\tan(θ + 28°) = \tan 30° − 1$$

Substituting \tan(30°) ≈ 0.5774 yields:

$$(1 + 0.5774)(\tan(θ + 28°) = 0.5774 − 1$$

This simplifies to:

$$(1.5774)(\tan(θ + 28°)) = -0.4226$$

Let's derive \tan(θ + 28°):

$$\tan(θ + 28°) = \frac{-0.4226}{1.5774} ≈ -0.268$$

Next, we find:

$$θ + 28° = \tan^{-1}(-0.268)$$

This gives us two solutions:

$$θ + 28° ≈ 180° - 15° + k imes 180° ext{ (where k ∈ ℤ)}$$

Solving for θ in the valid range gives:

$$θ ≈ -13° ext{ or } θ ≈ 126.5°$$

The only valid solution for 0 < θ < 180° is θ ≈ 126.5°.