A-Level AQA Maths Pure 5.8 Trigonometric Proof: Prove the identity $$\cot^2 \t

Prove the identity $$\cot^2 \theta - \cos^2 \theta = \cot^2 \theta \cos^2 \theta$$ - AQA - A-Level Maths Pure - Question 13 - 2017 - Paper 1

Question 13

$Prove-the-identity---$$\cot^2-\theta---\cos^2-\theta-=-\cot^2-\theta-\cos^2-\theta$$-AQA-A-Level Maths Pure-Question 13-2017-Paper 1.png$

Prove the identity $$\cot^2 \theta - \cos^2 \theta = \cot^2 \theta \cos^2 \theta$$

Worked Solution & Example Answer:Prove the identity $$\cot^2 \theta - \cos^2 \theta = \cot^2 \theta \cos^2 \theta$$ - AQA - A-Level Maths Pure - Question 13 - 2017 - Paper 1

Step 1

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Answer

We start by writing the left-hand side of the equation:

$\cot^2 \theta - \cos^2 \theta$

We know that: $\cot \theta = \frac{\cos \theta}{\sin \theta}$ Thus, we have:

$\cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta}$

Substituting this into the left-hand side gives:

$\frac{\cos^2 \theta}{\sin^2 \theta} - \cos^2 \theta$

Step 2

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Answer

Now, we need a common denominator to combine the terms:

$LHS = \frac{\cos^2 \theta - \cos^2 \theta \sin^2 \theta}{\sin^2 \theta}$

Factoring out (\cos^2 \theta) from the numerator, we get:

$LHS = \frac{\cos^2 \theta (1 - \sin^2 \theta)}{\sin^2 \theta}$

Using the Pythagorean identity, we know that:

$1 - \sin^2 \theta = \cos^2 \theta$

Thus, we can substitute this back into our equation:

Step 3

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Answer

$LHS = \frac{\cos^2 \theta \cos^2 \theta}{\sin^2 \theta} = \frac{\cos^4 \theta}{\sin^2 \theta}$

We can express $\frac{1}{\sin^2 \theta}$ as $\csc^2 \theta$ :

$LHS = \cos^4 \theta \csc^2 \theta$

And since: $\cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta \cos^2 \theta$

This matches the right-hand side, confirming the identity.