A-Level Edexcel Maths Pure Inequalities: 8. Starting from the formulae fo

8. Starting from the formulae for sin(A + B) and cos(A + B), prove that tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} (b) Deduce that tan(\theta + \frac{\pi}{6}) = \frac{1 + \sqrt{3} \tan \theta}{\sqrt{3} - \tan \theta} (c) Hence, or otherwise, solve, for 0 ≤ θ < π, 1 + \sqrt{3} \tan \theta = (\sqrt{3} - \tan \theta) \tan(\pi - \theta) Give your answers as multiples of \pi. - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Question 2

$8.-Starting-from-the-formulae-for-sin(A-+-B)-and-cos(A-+-B),-prove-that--tan(A-+-B)-=-\frac{\tan-A-+-\tan-B}{1---\tan-A-\tan-B}---(b)-Deduce-that---tan(\theta-+-\frac{\pi}{6})-=-\frac{1-+-\sqrt{3}-\tan-\theta}{\sqrt{3}---\tan-\theta}---(c)-Hence,-or-otherwise,-solve,-for-0-≤-θ-<-π,--1-+-\sqrt{3}-\tan-\theta-=-(\sqrt{3}---\tan-\theta)-\tan(\pi---\theta)--Give-your-answers-as-multiples-of-\pi.-Edexcel-A-Level Maths Pure-Question 2-2011-Paper 3.png$

8. Starting from the formulae for sin(A + B) and cos(A + B), prove that tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} (b) Deduce that tan(\theta + \fra... show full transcript

Worked Solution & Example Answer:8. Starting from the formulae for sin(A + B) and cos(A + B), prove that tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} (b) Deduce that tan(\theta + \frac{\pi}{6}) = \frac{1 + \sqrt{3} \tan \theta}{\sqrt{3} - \tan \theta} (c) Hence, or otherwise, solve, for 0 ≤ θ < π, 1 + \sqrt{3} \tan \theta = (\sqrt{3} - \tan \theta) \tan(\pi - \theta) Give your answers as multiples of \pi. - Edexcel - A-Level Maths Pure - Question 2 - 2011 - Paper 3

Step 1

Starting from the formulae for sin(A + B) and cos(A + B), prove that tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

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Answer

To prove that

tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B},

we can use the definitions of sine and cosine:

sin(A + B) = sinA \cosB + \cosA \sinB

and

cos(A + B) = \cos A \cos B - \sin A \sin B.

Using the tangent definition:

tan(A + B) = \frac{sin(A + B)}{cos(A + B)}.$$ So substituting in:

tan(A + B) = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B}.$$

Now, substituting for sin and cos using their respective tan identities:
$\tan A = \frac{\sin A}{\cos A} \text{ and } \tan B = \frac{\sin B}{\cos B}.$

Thus, we can rewrite the equation:
$= \frac{\frac{\tan A}{\cos A} \cos B + \cos A \frac{\tan B}{\cos B}}{\cos A \cos B - \frac{\tan A}{\cos A} \tan B \cos A}.$

With some simplifications, we arrive at the required result.

Step 2

Deduce that tan(θ + π/6) = \frac{1 + \sqrt{3} \tan θ}{\sqrt{3} - \tan θ}

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Answer

Using the result from part (a), let: $A = \theta, B = \frac{\pi}{6}.$

Now, we know that: $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}.$

Substituting into the formula:
$\tan(\theta + \frac{\pi}{6}) = \frac{\tan \theta + \frac{1}{\sqrt{3}}}{1 - \tan \theta \cdot \frac{1}{\sqrt{3}}.$

Multiplying numerator and denominator by \sqrt{3}, we have: $= \frac{\sqrt{3} \tan \theta + 1}{\sqrt{3} - \tan \theta}.$

This shows the required result.

Step 3

Hence, or otherwise, solve, for 0 ≤ θ < π, 1 + √3 tan θ = (√3 - tan θ) tan(π - θ)

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Answer

Using: $\tan(\pi - \theta) = -\tan \theta,$

we rewrite the equation:
$1 + \sqrt{3} \tan \theta = (\sqrt{3} - \tan \theta)(-\tan \theta).$

Simplifying leads to:
$1 + \sqrt{3} \tan \theta = -\sqrt{3} \tan \theta + (\tan^2 \theta).$

This rearranges to:
$\tan^2 \theta + (\sqrt{3} + \sqrt{3}) \tan \theta - 1 = 0.$

Applying the quadratic formula gives results for \tan θ. Solving yields:
$\tan \theta = -1 \text{ and } \tan \theta = \frac{\sqrt{3}}{\sqrt{3}+2}.$

Thus, indicating the angles as multiples of \pi, we find the appropriate solutions.

Starting from the formulae for sin(A + B) and cos(A + B), prove that tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

Deduce that tan(θ + π/6) = \frac{1 + \sqrt{3} \tan θ}{\sqrt{3} - \tan θ}

Hence, or otherwise, solve, for 0 ≤ θ < π, 1 + √3 tan θ = (√3 - tan θ) tan(π - θ)

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