<p>If \(\cos(x-y)\), \(\cos x\) and \(\cos(x+y)\) are in HP then \(\cos x \cdot \sec\left(\frac{y}{2}\right) =\) ______</p>
Step-by-Step Solution
Key Concept: If three terms are in HP, their reciprocals are in AP. Use this to convert the HP condition into an AP condition for reciprocals, then apply trigonometric identities for cos(x±y).
<p><strong>Step 1:</strong> If cos(x−y), cos x, cos(x+y) are in HP, then their reciprocals are in AP.</p><p>So: 2/cos x = 1/cos(x−y) + 1/cos(x+y)</p><p><strong>Step 2:</strong> Simplify the right side:</p><p>2/cos x = [cos(x+y) + cos(x−y)] / [cos(x−y)·cos(x+y)]</p><p><strong>Step 3:</strong> Use sum-to-product formula: cos(x+y) + cos(x−y) = 2cos x cos y</p><p>2/cos x = 2cos x cos y / [cos(x−y)·cos(x+y)]</p><p><strong>Step 4:</strong> Cross multiply: 2cos(x−y)·cos(x+y) = 2cos²x cos y</p><p>cos(x−y)·cos(x+y) = cos²x cos y</p><p><strong>Step 5:</strong> Use product-to-sum: cos(x−y)·cos(x+y) = (1/2)[cos 2x + cos 2y]</p><p>(1/2)[cos 2x + cos 2y] = cos²x cos y</p><p><strong>Step 6:</strong> Substitute cos 2x = 2cos²x − 1 and cos 2y = 2cos²y − 1:</p><p>(1/2)[2cos²x − 1 + 2cos²y − 1] = cos²x cos y</p><p>cos²x − 1 + cos²y = cos²x cos y</p><p>cos²y − 1 = cos²x(cos y − 1)</p><p>−sin²y = −cos²x(1 − cos y)</p><p><strong>Step 7:</strong> Using 1 − cos y = 2sin²(y/2) and sin²y = 4sin²(y/2)cos²(y/2):</p><p>4sin²(y/2)cos²(y/2) = 2cos²x·sin²(y/2)</p><p>2cos²(y/2) = cos²x</p><p>cos x·sec(y/2) = ±√2</p><p>∴ Answer: <strong>±√2</strong></p>
Correct Answer: ±√2