Trigonometry & Inverse Trigonometry
Trigonometric Identities and Equations
Grade 11
Question:
<p>The value of <span class="math">\frac{2\cos^3\left(\frac{\pi}{2}+x\right)\cot(3\pi+x)\sec(x-3\pi)\operatorname{cosec}\left(\frac{3\pi}{2}-x\right)}{\cot x\tan^2(x-\pi)\sin(x-2\pi)}\)</span> is equal to</p>
<p>(a) 0</p>
<p>(b) -1</p>
<p>(c) -2</p>
<p>(d) 1</p>
Step-by-Step Solution
Key Concept: Use reduction formulas to simplify trigonometric expressions with compound angles, then algebraically reduce the fraction.
<p><strong>Step 1:</strong> Simplify using trigonometric identities for compound angles.</p><p><strong>Step 2:</strong> Apply reduction formulas:</p><ul><li>\(\cos\left(\frac{\pi}{2}+x\right) = -\sin x\)</li><li>\(\cot(3\pi+x) = \cot x\)</li><li>\(\sec(x-3\pi) = -\sec x\)</li><li>\(\operatorname{cosec}\left(\frac{3\pi}{2}-x\right) = -\sec x\)</li><li>\(\tan^2(x-\pi) = \tan^2 x\)</li><li>\(\sin(x-2\pi) = \sin x\)</li></ul><p><strong>Step 3:</strong> \(\frac{2(-\sin x)^3(\cot x)(-\sec x)(-\sec x)}{\cot x \tan^2 x \sin x} = \frac{2(-\sin x)^3 \cdot (-\sec^2 x)}{\tan^2 x \sin x} = -2\)</p><p>∴ Answer is (c) -2</p>
Correct Answer: C