Trigonometry & Inverse Trigonometry
Trigonometric Identities
Grade 11
Question:
<p>The value of \(\dfrac{1 - \tan^2 15°}{1 + \tan^2 15°}\) is</p>
<p>1</p>
<p>\(\sqrt{3}\)</p>
<p>\(\dfrac{\sqrt{3}}{2}\)</p>
<p>2</p>
Step-by-Step Solution
Key Concept: Recognize that the expression (1 - tan²θ)/(1 + tan²θ) is the double angle formula for cosine: cos(2θ). Simply substitute θ = 15° to get cos(30°).
<p><strong>Step 1:</strong> Recognize the trigonometric identity.</p><p>The expression $\frac{1 - \tan^2\theta}{1 + \tan^2\theta}$ is the double angle formula for cosine:</p><p>$$\cos(2\theta) = \frac{1 - \tan^2\theta}{1 + \tan^2\theta}$$</p><p><strong>Step 2:</strong> Apply the identity with θ = 15°.</p><p>$$\frac{1 - \tan^2 15°}{1 + \tan^2 15°} = \cos(2 \times 15°) = \cos(30°)$$</p><p><strong>Step 3:</strong> Evaluate cos(30°).</p><p>$$\cos(30°) = \frac{\sqrt{3}}{2}$$</p><p>∴ Answer: <strong>C</strong> (which is $\frac{\sqrt{3}}{2}$)</p>
Correct Answer: C