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Difficulty: HardBasic Trigonometric Ratios, Special Angles, and Identities

Given that sin⁡θ+cos⁡θ=62\sin \theta + \cos \theta = \frac{\sqrt{6}}{2} where θ\theta is an acute angle, what is the exact value of tan⁡θ+cot⁡θ\tan \theta + \cot \theta?

  1. 4Answer
  2. B
    2
  3. C
    23\frac{2}{3}
  4. D
    12\frac{1}{2}

Answer

The exact value of tan⁡θ+cot⁡θ\tan \theta + \cot \theta is 4.
By squaring both sides of sin⁡θ+cos⁡θ=62\sin \theta + \cos \theta = \frac{\sqrt{6}}{2}, we obtain sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=64=32\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = \frac{6}{4} = \frac{3}{2}. Applying the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 simplifies this to 1+2sin⁡θcos⁡θ=321 + 2\sin \theta \cos \theta = \frac{3}{2}, which yields sin⁡θcos⁡θ=14\sin \theta \cos \theta = \frac{1}{4}. Since tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}, the exact value is 11/4=4\frac{1}{1/4} = 4.

Step-by-Step Solution

1
Square both sides of the given equation sin⁡θ+cos⁡θ=62\sin \theta + \cos \theta = \frac{\sqrt{6}}{2}.
(sin⁡θ+cos⁡θ)2=(62)2  ⟹  sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=64=32(\sin \theta + \cos \theta)^2 = \left(\frac{\sqrt{6}}{2}\right)^2 \implies \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = \frac{6}{4} = \frac{3}{2}.
Squaring allows the application of the fundamental Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to isolate the product sin⁡θcos⁡θ\sin \theta \cos \theta.
2
Substitute sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 into the equation and solve for sin⁡θcos⁡θ\sin \theta \cos \theta.
1+2sin⁡θcos⁡θ=32  ⟹  2sin⁡θcos⁡θ=12  ⟹  sin⁡θcos⁡θ=141 + 2\sin \theta \cos \theta = \frac{3}{2} \implies 2\sin \theta \cos \theta = \frac{1}{2} \implies \sin \theta \cos \theta = \frac{1}{4}.
Simplifying the algebraic equation isolates the product term.
3
Express tan⁡θ+cot⁡θ\tan \theta + \cot \theta in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta.
tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}.
Using basic trigonometric identities simplifies the target sum into a reciprocal product.
4
Substitute sin⁡θcos⁡θ=14\sin \theta \cos \theta = \frac{1}{4} into the simplified expression.
tan⁡θ+cot⁡θ=11/4=4\tan \theta + \cot \theta = \frac{1}{1/4} = 4.
Evaluating the reciprocal fraction gives the final numerical answer.

Key Concept

Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 and reciprocal relation tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ\tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta}.
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Topics

Basic Trigonometric Ratios, Special Angles, and Identities

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