Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>The value of <span class='mathjax'>\(\tan^{-1} \left[ \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right]\)</span> is</p>
<p>(a) \(\frac{-x}{4(1-x^2)}\)</p>
<p>(b) \(\frac{x}{4(1-x^2)}\)</p>
<p>(c) \(\frac{x}{2(1-x^2)}\)</p>
<p>(d) \(\frac{-x}{2(1-x^2)}\)</p>
Step-by-Step Solution
Key Concept: Use substitution $x^2 = \cos 2\theta$ and trigonometric identities to simplify inverse trigonometric expressions.
<p><strong>Solution:</strong> Assume $x^2 = \cos 2\theta$ to simplify and then use the relations $1 + \cos 2\theta = 2\cos^2 \theta$ and $1 - \cos 2\theta = 2\sin^2 \theta$ to simplify the expression.</p><p>On putting $x^2 = \cos 2\theta$, we get:</p><p>$y = \tan^{-1} \left( \frac{\sqrt{1 + \cos 2\theta} + \sqrt{1 - \cos 2\theta}}{\sqrt{1 + \cos 2\theta} - \sqrt{1 - \cos 2\theta}} \right)$</p><p>$= \tan^{-1} \left( \frac{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta} \right)$</p><p>$= \tan^{-1} \left( \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \right)$</p><p>$= \tan^{-1} \left( \frac{1 + \tan\theta}{1 - \tan\theta} \right)$</p><p>$= \tan^{-1} \left[ \tan\left(\frac{\pi}{4} + \theta\right) \right]$</p><p>∴ Answer is (b).</p>
Correct Answer: B