Trigonometry & Inverse Trigonometry
Trigonometric equations
Grade 11

Question:

<p>If <span class="math">\frac{1+\sin A}{\sqrt{1-\sin A}} - \sqrt{1-\sin A} = 2\cos\frac{A}{2}\)</span>, then value of <span class="math">A\)</span> can be</p>
<p>(a) 110°</p>
<p>(b) 260°</p>
<p>(c) 300°</p>
<p>(d) 190°</p>

Step-by-Step Solution

Key Concept: Use the identities for sum and difference of square roots involving sine and cosine half-angles to verify which angles satisfy the equation.
<p><strong>Solution:</strong> Starting with <span class="math">\sqrt{1+\sin A} - \sqrt{1-\sin A} = 2\cos\frac{A}{2}\)</span></p><p>We can write: <span class="math">\sin\frac{A}{2} + \cos\frac{A}{2} - \sin\frac{A}{2} + \cos\frac{A}{2} = 2\cos\frac{A}{2}\)</span></p><p>This simplifies to expressions involving <span class="math">\sin\frac{A}{2}\)</span> and <span class="math">\cos\frac{A}{2}\)</span></p><p>Testing the given options: <span class="math">A = 110°, 260°, 190°\)</span> satisfy the equation.</p><p>∴ Answer is (a), (b), (d).</p>
Correct Answer: a,b,d

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