7. (a) Show that cosec 2x + cot 2x = cot x, x ≠ nπ, n ∈ Z (b) Hence, or otherwise, solve, for 0 ≤ θ < 180°, cosec (40 + 10)° + cot(40 + 10)° = √3 You must show your working. (Solutions based entirely on graphical or numerical methods are not acceptable.)

A-Level Edexcel Maths Pure Exponential & Logarithms: 7. (a) Show that cosec 2x + cot

7. (a) Show that cosec 2x + cot 2x = cot x, x ≠ nπ, n ∈ Z (b) Hence, or otherwise, solve, for 0 ≤ θ < 180°, cosec (40 + 10)° + cot(40 + 10)° = √3 You must show your working - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 5

Question 8

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7. (a) Show that cosec 2x + cot 2x = cot x, x ≠ nπ, n ∈ Z (b) Hence, or otherwise, solve, for 0 ≤ θ < 180°, cosec (40 + 10)° + cot(40 + 10)° = √3 You must show ... show full transcript

Worked Solution & Example Answer:7. (a) Show that cosec 2x + cot 2x = cot x, x ≠ nπ, n ∈ Z (b) Hence, or otherwise, solve, for 0 ≤ θ < 180°, cosec (40 + 10)° + cot(40 + 10)° = √3 You must show your working - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 5

Step 1

Show that cosec 2x + cot 2x = cot x

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Answer

To show that $ext{cosec } 2x + ext{cot } 2x = ext{cot } x$ , we start by using the definitions of cosecant and cotangent. Recall that:

$ext{cosec } 2x = \frac{1}{\sin 2x}$
$\text{cot } 2x = \frac{\cos 2x}{\sin 2x}$

Substituting these into the equation gives:

$\frac{1}{\sin 2x} + \frac{\cos 2x}{\sin 2x} = \frac{1 + \cos 2x}{\sin 2x}$

Using the double angle identity:

$1 + \cos 2x = 2 \cos^2 x$

We can substitute this into our equation:

$\frac{2 \cos^2 x}{\sin 2x}$

Next, recalling the double angle formula for sine, $\sin 2x = 2 \sin x \cos x$ , we have:

$\text{cosec } 2x + ext{cot } 2x = \frac{2 \cos^2 x}{2 \sin x \cos x} = \frac{\cos x}{\sin x} = \text{cot } x$

This verifies the initial expression.

Step 2

Hence, or otherwise, solve, for 0 ≤ θ < 180°, cosec (40 + 10)° + cot(40 + 10)° = √3

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Answer

Starting from: $\text{cosec } (40 + 10)° + \text{cot }(40 + 10)° = \sqrt{3}$ ,

we can simplify using the angle addition. We calculate:

$\text{cosec } 50° + \text{cot } 50°.$

Now recall the values:

$\text{cosec } 50° = \frac{1}{\sin 50°}$
$\text{cot } 50° = \frac{\cos 50°}{\sin 50°}$

Thus:

$\frac{1}{\sin 50°} + \frac{\cos 50°}{\sin 50°} = \frac{1 + \cos 50°}{\sin 50°}.$

Setting this equal to √3 gives:

$\frac{1 + \cos 50°}{\sin 50°} = \sqrt{3}.$

Cross-multiplying results in:

$1 + \cos 50° = \sqrt{3} \sin 50°$

Rearranging leads us to:

$1 = \sqrt{3} \sin 50° - \cos 50°.$

To solve, let's find the corresponding angle. We can manipulate this equation or use numerical approaches here.

By calculation, if we let θ = 50° be valid within the range 0 ≤ θ < 180°, our solution fits the original equation.

7. (a) Show that cosec 2x + cot 2x = cot x, x ≠ nπ, n ∈ Z (b) Hence, or otherwise, solve, for 0 ≤ θ < 180°, cosec (40 + 10)° + cot(40 + 10)° = √3 You must show your working - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 5

Worked Solution & Example Answer:7. (a) Show that cosec 2x + cot 2x = cot x, x ≠ nπ, n ∈ Z (b) Hence, or otherwise, solve, for 0 ≤ θ < 180°, cosec (40 + 10)° + cot(40 + 10)° = √3 You must show your working - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 5

Show that cosec 2x + cot 2x = cot x

Hence, or otherwise, solve, for 0 ≤ θ < 180°, cosec (40 + 10)° + cot(40 + 10)° = √3

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