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Difficulty: Very hardFundamental Trigonometric Identities

An angle θ\theta satisfies π2<θ<π\frac{\pi}{2} < \theta < \pi and (sin⁡θ−cos⁡θ)2=179(\sin\theta - \cos\theta)^2 = \frac{17}{9}. What is the value of tan⁡θ+cot⁡θ\tan\theta + \cot\theta?

  1. A
    11
  2. B
    −98-\frac{9}{8}
  3. −94-\frac{9}{4}Answer
  4. D
    94\frac{9}{4}
  5. E
    −49-\frac{4}{9}

Answer

The value of the expression is −94-\frac{9}{4}
Expanding the square of the difference (sin⁡θ−cos⁡θ)2(\sin\theta - \cos\theta)^2 yields sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta. Applying the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 simplifies this expression to 1−2sin⁡θcos⁡θ1 - 2\sin\theta\cos\theta. Setting this equal to 179\frac{17}{9} and solving for the product of sine and cosine gives sin⁡θcos⁡θ=−49\sin\theta\cos\theta = -\frac{4}{9}. The target expression tan⁡θ+cot⁡θ\tan\theta + \cot\theta can be rewritten using quotient and reciprocal identities as sin⁡θcos⁡θ+cos⁡θsin⁡θ\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}, which simplifies by finding a common denominator to 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}. Substituting the value of sin⁡θcos⁡θ\sin\theta\cos\theta into this expression results in −94-\frac{9}{4}.

Step-by-Step Solution

1
Expand the squared expression (sin⁡θ−cos⁡θ)2(\sin\theta - \cos\theta)^2 and apply the Pythagorean identity.
(sin⁡θ−cos⁡θ)2=sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=1−2sin⁡θcos⁡θ(\sin\theta - \cos\theta)^2 = \sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 1 - 2\sin\theta\cos\theta
To express the squared binomial in terms of the product sin⁡θcos⁡θ\sin\theta\cos\theta using the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
2
Equate the expanded form to the given value and solve for sin⁡θcos⁡θ\sin\theta\cos\theta.
1−2sin⁡θcos⁡θ=179  ⟹  2sin⁡θcos⁡θ=−89  ⟹  sin⁡θcos⁡θ=−491 - 2\sin\theta\cos\theta = \frac{17}{9} \implies 2\sin\theta\cos\theta = -\frac{8}{9} \implies \sin\theta\cos\theta = -\frac{4}{9}
To find the value of the product sin⁡θcos⁡θ\sin\theta\cos\theta from the given equation.
3
Rewrite the target expression tan⁡θ+cot⁡θ\tan\theta + \cot\theta in terms of sine and cosine.
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}
To simplify the target sum of ratios using quotient and reciprocal identities so that it depends only on the product sin⁡θcos⁡θ\sin\theta\cos\theta.
4
Substitute the value of sin⁡θcos⁡θ\sin\theta\cos\theta into the simplified expression.
tan⁡θ+cot⁡θ=1−4/9=−94\tan\theta + \cot\theta = \frac{1}{-4/9} = -\frac{9}{4}
To compute the final numerical value of the expression.

Key Concept

Simplifying trigonometric expressions using fundamental Pythagorean, quotient, and reciprocal identities, and solving for unknown products of trigonometric functions.

Alternative Method

Alternatively, one can find the individual values of sin⁡θ\sin\theta and cos⁡θ\cos\theta. Since (sin⁡θ−cos⁡θ)2=179(\sin\theta - \cos\theta)^2 = \frac{17}{9} and sin⁡θcos⁡θ=−49\sin\theta\cos\theta = -\frac{4}{9}, we can use the identity (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ=19(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta = \frac{1}{9}. In Quadrant II, sin⁡θ>0\sin\theta > 0 and cos⁡θ<0\cos\theta < 0, and since −cos⁡θ>0-\cos\theta > 0, we have sin⁡θ−cos⁡θ=173\sin\theta - \cos\theta = \frac{\sqrt{17}}{3}. Solving the system of equations for sin⁡θ\sin\theta and cos⁡θ\cos\theta yields sin⁡θ=17−16\sin\theta = \frac{\sqrt{17} - 1}{6} (since it must be positive) and cos⁡θ=−17−16\cos\theta = \frac{-\sqrt{17} - 1}{6} (since it must be negative). Substituting these into tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} will yield the same result of −94-\frac{9}{4}, though this method involves significantly more algebraic work.
Estimated Time:3m 0s
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Topics

Fundamental Trigonometric Identities

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