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Difficulty: MediumFundamental Trigonometric Identities

If sin⁡θ+cos⁡θ=1.5\sin \theta + \cos \theta = \sqrt{1.5}, what is the value of tan⁡θ+cot⁡θ\tan \theta + \cot \theta?

Answer: 4

Answer

4
Squaring both sides of sin⁡θ+cos⁡θ=1.5\sin \theta + \cos \theta = \sqrt{1.5} yields sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=1.5\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5. Using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, we get 1+2sin⁡θcos⁡θ=1.51 + 2\sin \theta \cos \theta = 1.5, which simplifies to sin⁡θcos⁡θ=0.25\sin \theta \cos \theta = 0.25. Rewriting tan⁡θ+cot⁡θ\tan \theta + \cot \theta using quotient identities gives sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\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}. Substituting sin⁡θcos⁡θ=0.25\sin \theta \cos \theta = 0.25 yields 10.25=4\frac{1}{0.25} = 4.

Step-by-Step Solution

1
Square both sides of the given equation
sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=1.5\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5
Squaring both sides allows the expansion of the binomial (sin⁡θ+cos⁡θ)2(\sin \theta + \cos \theta)^2 to reveal the product term sin⁡θcos⁡θ\sin \theta \cos \theta.
2
Apply the Pythagorean identity to solve for sin⁡θcos⁡θ\sin \theta \cos \theta
sin⁡θcos⁡θ=0.25\sin \theta \cos \theta = 0.25
Since sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, substituting 11 into 1+2sin⁡θcos⁡θ=1.51 + 2\sin \theta \cos \theta = 1.5 gives 2sin⁡θcos⁡θ=0.52\sin \theta \cos \theta = 0.5, so sin⁡θcos⁡θ=0.25\sin \theta \cos \theta = 0.25.
3
Rewrite tan⁡θ+cot⁡θ\tan \theta + \cot \theta using quotient identities
tan⁡θ+cot⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}
Using tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} and cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}, finding a common denominator yields sin⁡2θ+cos⁡2θsin⁡θcos⁡θ\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}. Applying sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 simplifies the numerator to 11.
4
Substitute the numerical value of sin⁡θcos⁡θ\sin \theta \cos \theta to find the answer
tan⁡θ+cot⁡θ=10.25=4\tan \theta + \cot \theta = \frac{1}{0.25} = 4
Dividing 11 by 0.250.25 evaluates to the exact integer 44.

Key Concept

Fundamental Trigonometric Identities (Pythagorean and Quotient Identities)
Estimated Time:1m 30s
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Topics

Fundamental Trigonometric Identities

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