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Difficulty: MediumFundamental Trigonometric Identities

Which of the following expressions is equivalent to tan⁡θ+cot⁡θcsc⁡θ\frac{\tan \theta + \cot \theta}{\csc \theta} for all values of θ\theta where the expression is defined?

  1. sec⁡θ\sec \thetaAnswer
  2. B
    cos⁡θ\cos \theta
  3. C
    sin⁡θ\sin \theta
  4. D
    csc⁡θ\csc \theta
  5. E
    tan⁡2θ\tan^2 \theta

Answer

sec⁡θ\sec \theta
Substituting 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} gives sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} in the numerator. Dividing by csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta} cancels the sin⁡θ\sin \theta term, leaving 1cos⁡θ\frac{1}{\cos \theta}, which is equal to sec⁡θ\sec \theta.

Step-by-Step Solution

1
Express tangent and cotangent using sine and cosine
tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}
Quotient identities allow rewriting all functions in terms of sine and cosine.
2
Combine the fractions in the numerator over a common denominator
\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}
Finding a common denominator of sin⁡θcos⁡θ\sin \theta \cos \theta allows application of the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.
3
Divide by the denominator csc⁡θ\csc \theta
\frac{\frac{1}{\sin \theta \cos \theta}}{\frac{1}{\sin \theta}} = \frac{1}{\sin \theta \cos \theta} \cdot \frac{\sin \theta}{1} = \frac{1}{\cos \theta} = \sec \theta
The cosecant function is the reciprocal of sine, so dividing by csc⁡θ\csc \theta cancels out the sin⁡θ\sin \theta factor in the denominator.

Key Concept

Fundamental Trigonometric Identities
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Topics

Fundamental Trigonometric Identities

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