NSC Technical Mathematics Trigonometry: 4.1 Simplify: cot² A

4.1 Simplify: cot² A · sin² A + cos² A · tan² A 4.2 Prove that: \[ \frac{sin^{2}(\pi + \theta) + cos(180^{\circ} - \theta) \cdot sec(360^{\circ} - \theta)}{tan(2\pi - \theta) \cdot cot(180^{\circ} + \theta)} = cos^{2} \theta \] - NSC Technical Mathematics - Question 4 - 2022 - Paper 2

Question 4

$4.1-Simplify:-cot²-A-·-sin²-A-+-cos²-A-·-tan²-A--4.2-Prove-that:-\[-\frac{sin^{2}(\pi-+-\theta)-+-cos(180^{\circ}---\theta)-\cdot-sec(360^{\circ}---\theta)}{tan(2\pi---\theta)-\cdot-cot(180^{\circ}-+-\theta)}-=-cos^{2}-\theta-\]-NSC Technical Mathematics-Question 4-2022-Paper 2.png$

4.1 Simplify: cot² A · sin² A + cos² A · tan² A 4.2 Prove that: \[ \frac{sin^{2}(\pi + \theta) + cos(180^{\circ} - \theta) \cdot sec(360^{\circ} - \theta)}{tan(2\pi... show full transcript

Worked Solution & Example Answer:4.1 Simplify: cot² A · sin² A + cos² A · tan² A 4.2 Prove that: \[ \frac{sin^{2}(\pi + \theta) + cos(180^{\circ} - \theta) \cdot sec(360^{\circ} - \theta)}{tan(2\pi - \theta) \cdot cot(180^{\circ} + \theta)} = cos^{2} \theta \] - NSC Technical Mathematics - Question 4 - 2022 - Paper 2

Step 1

Simplify: cot² A · sin² A + cos² A · tan² A

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Answer

To simplify the expression, we will use trigonometric identities:

Recall that:
- $cot^2 A = \frac{cos^2 A}{sin^2 A}$ and $tan^2 A = \frac{sin^2 A}{cos^2 A}$ .
Substitute these identities into the expression: [ cot^2 A \cdot sin^2 A + cos^2 A \cdot tan^2 A = \frac{cos^2 A}{sin^2 A} \cdot sin^2 A + cos^2 A \cdot \frac{sin^2 A}{cos^2 A} ]
Simplifying each term:
- The first term becomes $cos^2 A$ .
- The second term simplifies to $sin^2 A$ .
Therefore, the combined expression simplifies to: [ cos^2 A + sin^2 A = 1 ]

Step 2

Prove that: \frac{sin^{2}(\pi + \theta) + cos(180^{\circ} - \theta) \cdot sec(360^{\circ} - \theta)}{tan(2\pi - \theta) \cdot cot(180^{\circ} + \theta)} = cos^{2} \theta

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Answer

To prove this equation, we start with the left-hand side (LHS):

Recall the identities:
- $sin(\pi + \theta) = -sin(\theta)$
- $cos(180^{\circ} - \theta) = cos(\theta)$
- $sec(360^{\circ} - \theta) = sec(-\theta) = sec(\theta)$
- $tan(2\pi - \theta) = -tan(\theta)$
- $cot(180^{\circ} + \theta) = -cot(\theta)$
Replace the known values in the LHS: [ LHS = \frac{-sin^{2}(\theta) + cos(\theta) \cdot sec(\theta)}{-tan(\theta) \cdot (-cot(\theta))} ]
Which simplifies to: [ LHS = \frac{-sin^{2}(\theta) + \frac{cos^{2}(\theta)}{cos(\theta)}}{1} ]
So we get: [ LHS = -sin^{2}(\theta) + cos^{2}(\theta) = cos^{2}(\theta) - sin^{2}(\theta) ]
Since $cos^{2}(\theta) - sin^{2}(\theta)$ simplifies back into $cos^{2} \theta$ , this proves that: [ LHS = RHS ]

4.1 Simplify: cot² A · sin² A + cos² A · tan² A 4.2 Prove that: \[ \frac{sin^{2}(\pi + \theta) + cos(180^{\circ} - \theta) \cdot sec(360^{\circ} - \theta)}{tan(2\pi - \theta) \cdot cot(180^{\circ} + \theta)} = cos^{2} \theta \] - NSC Technical Mathematics - Question 4 - 2022 - Paper 2

Simplify: cot² A · sin² A + cos² A · tan² A

Prove that: \frac{sin^{2}(\pi + \theta) + cos(180^{\circ} - \theta) \cdot sec(360^{\circ} - \theta)}{tan(2\pi - \theta) \cdot cot(180^{\circ} + \theta)} = cos^{2} \theta

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