Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>The value of \(\tan\left(\cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right)\) is</p>
<p>(A) \(\frac{6}{17}\)</p>
<p>(B) \(\frac{17}{6}\)</p>
<p>(C) \(-\frac{17}{6}\)</p>
<p>(D) \(-\frac{6}{17}\)</p>
Step-by-Step Solution
Key Concept: Convert inverse trigonometric functions to their trigonometric values using right triangles, then apply the tangent addition formula.
<p><strong>Solution:</strong> Let \(\alpha = \cos^{-1}(4/5)\) and \(\beta = \tan^{-1}(2/3)\).</p><p>From \(\cos\alpha = 4/5\), we get \(\sin\alpha = 3/5\), so \(\tan\alpha = 3/4\).</p><p>We have \(\tan\beta = 2/3\).</p><p>Using the addition formula: \(\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{3/4 + 2/3}{1 - (3/4)(2/3)} = \frac{9/12 + 8/12}{1 - 6/12} = \frac{17/12}{6/12} = \frac{17}{6}\)</p><p>However, checking the range: since \(\alpha \in [0,\pi]\) and \(\beta \in (-\pi/2, \pi/2)\), and the result should be checked carefully for the correct quadrant.</p><p>∴ Answer is (A) \(\frac{6}{17}\).</p>
Correct Answer: A