Sequences & Series
Arithmetic Progression - Trigonometric Application
Grade 11
Question:
<p>If \(q_1, q_2, q_3, \ldots, q_n\) are in AP, whose common difference is d, then \(\sin d (\sec q_1 \sec q_2 + \sec q_2 \sec q_3 + \ldots + \sec q_{n-1} \sec q_n)\) is equal to</p>
<p>(a) \(\tan q_n - \tan q_1\)</p>
<p>(b) \(\tan q_n + \tan q_1\)</p>
<p>(c) \(\tan q_n - \tan q_2\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the trigonometric identity that tan(A - B) = (tan A - tan B)/(1 + tan A tan B), and recognize that sec A sec B = 1/(cos A cos B). The key is to express sec q_i sec q_{i+1} in terms of tangent differences using the angle difference d.
<p><strong>Step 1: Recognize the identity to use</strong></p><p>Since q_i are in AP with common difference d, we have q_{i+1} = q_i + d. We use the identity:</p><p>tan(A) - tan(B) = sin(A - B)/(cos A cos B) = sin(A - B) · sec A sec B</p><p><strong>Step 2: Apply the identity with A = q_{i+1}, B = q_i</strong></p><p>For consecutive terms where q_{i+1} - q_i = d:</p><p>tan(q_{i+1}) - tan(q_i) = sin(d) · sec q_i sec q_{i+1}</p><p>Therefore: sin(d) · sec q_i sec q_{i+1} = tan(q_{i+1}) - tan(q_i)</p><p><strong>Step 3: Write the sum</strong></p><p>sin d(sec q_1 sec q_2 + sec q_2 sec q_3 + ... + sec q_{n-1} sec q_n)</p><p>= sin d · sec q_1 sec q_2 + sin d · sec q_2 sec q_3 + ... + sin d · sec q_{n-1} sec q_n</p><p><strong>Step 4: Apply the identity to each term</strong></p><p>= (tan q_2 - tan q_1) + (tan q_3 - tan q_2) + (tan q_4 - tan q_3) + ... + (tan q_n - tan q_{n-1})</p><p><strong>Step 5: Recognize the telescoping series</strong></p><p>This is a telescoping series where all intermediate terms cancel:</p><p>= -tan q_1 + tan q_n = tan q_n - tan q_1</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a