Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>If the mapping <math>f(x) = mx + c, m > 0</math> maps <math>[-1, 1]</math> onto <math>[0, 2]</math>, then <math>\tan\left(\tan^{-1}\frac{1}{7} + \cot^{-1}8 + \cot^{-1}18\right)</math> is equal to</p>
<p>(a) <math>\frac{1}{2}</math></p>
<p>(b) <math>\frac{3}{2}</math></p>
<p>(c) 0</p>
<p>(d) 1</p>
Step-by-Step Solution
Key Concept: Use the addition formula for inverse tangent functions and cotangent-to-tangent conversion to simplify the expression.
<p><strong>Solution:</strong></p><p>Clearly, <math>f(x) = x + 1</math> (As <math>-1 < x < 1 \Rightarrow 0 < x + 1 < 2</math>)</p><p>Now, <math>\tan\left(\tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8} + \tan^{-1}\frac{1}{18}\right)</math></p><p><math>= \tan\left(\tan^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\cdot\frac{1}{8}}\right) + \tan^{-1}\frac{1}{18}\right)</math></p><p><math>= \tan\left(\tan^{-1}\frac{15}{55} + \tan^{-1}\frac{1}{18}\right)</math></p><p><math>= 1</math></p>
Correct Answer: D