Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>If <span class="math">a\sin^{-1}x - b\cos^{-1}x = c</span>, then <span class="math">a\sin^{-1}x + b\cos^{-1}x</span> is equal to</p>
<p>(a) 0</p>
<p>(b) <span class="math">\frac{\pi ab + c(b-a)}{a+b}</span></p>
<p>(c) <span class="math">\frac{\pi ab + c(a-b)}{a+b}</span></p>
<p>(d) <span class="math">\frac{\pi ab - c(a+b)}{a+b}</span></p>
Step-by-Step Solution
Key Concept: Use the fundamental identity sin⁻¹x + cos⁻¹x = π/2 to relate the two inverse trigonometric expressions, then solve the system of two equations with two unknowns.
<p><strong>Step 1:</strong> Let sin⁻¹x = α and cos⁻¹x = β. We know the fundamental identity: α + β = π/2</p><p><strong>Step 2:</strong> From the given equation: aα - bβ = c ... (1)</p><p><strong>Step 3:</strong> We need to find: aα + bβ ... (2)</p><p><strong>Step 4:</strong> From the fundamental identity: β = π/2 - α, so we have two equations:</p><ul><li>aα - b(π/2 - α) = c</li><li>aα - bπ/2 + bα = c</li><li>(a + b)α = c + bπ/2 ... (equation A)</li></ul><p><strong>Step 5:</strong> Now add equation (1) and equation (2):</p><ul><li>(aα - bβ) + (aα + bβ) = c + (aα + bβ)</li><li>2aα = c + (aα + bβ)</li><li>aα + bβ = 2aα - c</li></ul><p><strong>Step 6:</strong> From equation A: α = (c + bπ/2)/(a + b)</p><p><strong>Step 7:</strong> Substitute back: aα + bβ = aα + b(π/2 - α) = aα + bπ/2 - bα = (a - b)α + bπ/2</p><p><strong>Step 8:</strong> Substituting α = (c + bπ/2)/(a + b):</p><ul><li>aα + bβ = (a - b) · (c + bπ/2)/(a + b) + bπ/2</li><li>= [(a - b)(c + bπ/2) + bπ/2(a + b)] / (a + b)</li><li>= [(a - b)c + (a - b)bπ/2 + abπ/2 + b²π/2] / (a + b)</li><li>= [(a - b)c + abπ/2 - b²π/2 + abπ/2 + b²π/2] / (a + b)</li><li>= [(a - b)c + abπ] / (a + b)</li><li>= [πab + c(a - b)] / (a + b)</li></ul><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B