Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>Let <i>x</i><sub>1</sub> and <i>x</i><sub>2</sub> (<i>x</i><sub>1</sub> > <i>x</i><sub>2</sub>) be roots of the equation \(\sin^{-1}(\cos(\tan^{-1}(\csc(\cot^{-1}x)))) = \frac{\pi}{6}\). Which of the following is/are correct?</p><p>(a) \(\sin^{-1}\frac{1}{x_1} + \cos^{-1}\frac{1}{x_2} = \pi\)</p><p>(b) \(\sin^{-1}\frac{1}{x_1} + \cos^{-1}\frac{1}{x_1} = 0\)</p><p>(c) \(\sin^{-1}\frac{1}{x_2} + \sin^{-1}\frac{1}{x_1} = 0\)</p><p>(d) \(\cos^{-1}\frac{1}{x_1} + \cos^{-1}\frac{1}{x_2} = \pi\)</p>
<p>(a) \(\sin^{-1}\frac{1}{x_1} + \cos^{-1}\frac{1}{x_2} = \pi\)</p>
<p>(b) \(\sin^{-1}\frac{1}{x_1} + \cos^{-1}\frac{1}{x_1} = 0\)</p>
<p>(c) \(\sin^{-1}\frac{1}{x_2} + \sin^{-1}\frac{1}{x_1} = 0\)</p>
<p>(d) \(\cos^{-1}\frac{1}{x_1} + \cos^{-1}\frac{1}{x_2} = \pi\)</p>
Step-by-Step Solution
Key Concept: Simplify the composition of inverse trigonometric functions using identities, then solve the resulting equation for the roots and verify each statement.
<p><strong>Step 1:</strong> Simplify $\sin^{-1}(\cos(\tan^{-1}(\csc(\cot^{-1}x)))) = \frac{\pi}{6}$</p><p><strong>Step 2:</strong> Note that $\csc(\cot^{-1}x) = \sqrt{1+x^2}$</p><p><strong>Step 3:</strong> Then $\sin^{-1}(\cos(\tan^{-1}\sqrt{1+x^2})) = \frac{\pi}{6}$</p><p><strong>Step 4:</strong> This simplifies to $\sin^{-1}\frac{1}{\sqrt{x^2+2}} = \frac{\pi}{6}$</p><p><strong>Step 5:</strong> Therefore $\frac{1}{\sqrt{x^2+2}} = \frac{1}{2}$</p><p><strong>Step 6:</strong> Solving: $x^2 + 2 = 4 \Rightarrow x^2 = 2$</p><p><strong>Step 7:</strong> Thus $x = \pm\sqrt{2}$, so $x_1 = \sqrt{2}$ and $x_2 = -\sqrt{2}$</p><p><strong>Step 8:</strong> Verify each option: (a), (c), and (d) are correct.</p>
Correct Answer: a,c,d