Trigonometry & Inverse Trigonometry
Trigonometric identities
Grade 11

Question:

<p>If ABCD is a cyclic quadrilateral then \(\cos A + \cos B + \cos C + \cos D\) is equal to</p>
<p>(a) 1</p>
<p>(b) -1</p>
<p>(c) 0</p>
<p>(d) none of these</p>

Step-by-Step Solution

Key Concept: In a cyclic quadrilateral, opposite angles are supplementary (sum to 180°), so cos C = -cos A and cos D = -cos B. This means paired opposite angles cancel out completely.
<p><strong>Step 1:</strong> Recall that in a cyclic quadrilateral ABCD, opposite angles are supplementary.</p><p>Therefore: ∠A + ∠C = 180° and ∠B + ∠D = 180°</p><p><strong>Step 2:</strong> From ∠A + ∠C = 180°, we get ∠C = 180° - ∠A</p><p>Therefore: cos C = cos(180° - A) = -cos A</p><p><strong>Step 3:</strong> Similarly, from ∠B + ∠D = 180°, we get ∠D = 180° - ∠B</p><p>Therefore: cos D = cos(180° - B) = -cos B</p><p><strong>Step 4:</strong> Now calculate the sum:</p><p>cos A + cos B + cos C + cos D = cos A + cos B + (-cos A) + (-cos B) = 0</p><p>∴ Answer: <strong>0</strong> (Option C)</p>
Correct Answer: C

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