A-Level Edexcel Maths Pure Trigonometric Functions: 7. (a) Show that $$\csc 2x + \c

7. (a) Show that $$\csc 2x + \cot 2x = \cot x, \quad x \neq n \cdot 90^\circ, \quad n \in \mathbb{Z}$$ (b) Hence, or otherwise, solve, for $0 \leq \theta < 180^\circ$, $$\csc (40 + 10^\circ) + \cot(40 + 10^\circ) = \sqrt{3}$$ You must show your working - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 5

Question 9

$7.-(a)-Show-that--$$\csc-2x-+-\cot-2x-=-\cot-x,-\quad-x-\neq-n-\cdot-90^\circ,-\quad-n-\in-\mathbb{Z}$$--(b)-Hence,-or-otherwise,-solve,-for-$0-\leq-\theta-<-180^\circ$,--$$\csc-(40-+-10^\circ)-+-\cot(40-+-10^\circ)-=-\sqrt{3}$$--You-must-show-your-working-Edexcel-A-Level Maths Pure-Question 9-2014-Paper 5.png$

7. (a) Show that $$\csc 2x + \cot 2x = \cot x, \quad x \neq n \cdot 90^\circ, \quad n \in \mathbb{Z}$$ (b) Hence, or otherwise, solve, for $0 \leq \theta < 180^\ci... show full transcript

Worked Solution & Example Answer:7. (a) Show that $$\csc 2x + \cot 2x = \cot x, \quad x \neq n \cdot 90^\circ, \quad n \in \mathbb{Z}$$ (b) Hence, or otherwise, solve, for $0 \leq \theta < 180^\circ$, $$\csc (40 + 10^\circ) + \cot(40 + 10^\circ) = \sqrt{3}$$ You must show your working - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 5

Step 1

Show that $ \csc 2x + \cot 2x = \cot x $

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Answer

To show that ( \csc 2x + \cot 2x = \cot x ), we start with the left-hand side:

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

Using the double angle identity, ( \cos 2x = 1 - 2\sin^2 x ), we can substitute it in:

$= \frac{1 + (1 - 2\sin^2 x)}{\sin 2x} = \frac{2 - 2\sin^2 x}{\sin 2x}$

Now, recall that ( \sin 2x = 2\sin x \cos x ):

$= \frac{2(1 - \sin^2 x)}{2\sin x \cos x} = \frac{2\cos^2 x}{2\sin x \cos x} = \frac{\cos x}{\sin x} = \cot x$ Thus, we have shown that ( \csc 2x + \cot 2x = \cot x ).

Step 2

Hence, or otherwise, solve, for $0 \leq \theta < 180^\circ$: $\csc(40 + 10^\circ) + \cot(40 + 10^\circ) = \sqrt{3}$

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Answer

To solve the equation:

$\csc(50^\circ) + \cot(50^\circ) = \sqrt{3},$

We first express ( \csc ) and ( \cot ) in terms of ( \sin ) and ( \cos ):

$\frac{1}{\sin(50^\circ)} + \frac{\cos(50^\circ)}{\sin(50^\circ)} = \sqrt{3}$

This simplifies to:

$\frac{1 + \cos(50^\circ)}{\sin(50^\circ)} = \sqrt{3}$

We can cross-multiply to obtain:

$1 + \cos(50^\circ) = \sqrt{3} \sin(50^\circ)$

Now, remember that: $\sin(50^\circ) = \sin(90^\circ - 40^\circ) = \cos(40^\circ)$

Substituting gives us:

$1 + \cos(50^\circ) = \sqrt{3} \cos(40^\circ)$

Now, we need to evaluate (\cos(50^\circ)) and solve for the angle. You may use a calculator or values from trigonometric tables. This would lead to the required solution.

Show that \( \csc 2x + \cot 2x = \cot x \)

Hence, or otherwise, solve, for \(0 \leq \theta < 180^\circ\): \(\csc(40 + 10^\circ) + \cot(40 + 10^\circ) = \sqrt{3}\)

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