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Difficulty: HardFundamental Trigonometric Identities

For an angle θ\theta such that π2<θ<π\frac{\pi}{2} < \theta < \pi, if cos⁡θ1−sin⁡θ=−3\frac{\cos\theta}{1 - \sin\theta} = -3, what is the value of sin⁡θ−tan⁡θ\sin\theta - \tan\theta?

  1. A
    −815-\frac{8}{15}
  2. B
    815\frac{8}{15}
  3. C
    11
  4. 3215\frac{32}{15}Answer
  5. E
    3512\frac{35}{12}

Answer

3215\frac{32}{15}
By multiplying the numerator and denominator of the given expression by 1+sin⁡θ1 + \sin\theta and utilizing the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, we find that sec⁡θ+tan⁡θ=−3\sec\theta + \tan\theta = -3. Since sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, it follows that sec⁡θ−tan⁡θ=−13\sec\theta - \tan\theta = -\frac{1}{3}. Solving this system of equations yields sec⁡θ=−53\sec\theta = -\frac{5}{3} (which means cos⁡θ=−35\cos\theta = -\frac{3}{5}) and tan⁡θ=−43\tan\theta = -\frac{4}{3}. In Quadrant II, sine is positive, so sin⁡θ=tan⁡θcos⁡θ=45\sin\theta = \tan\theta \cos\theta = \frac{4}{5}. Thus, sin⁡θ−tan⁡θ=45−(−43)=12+2015=3215\sin\theta - \tan\theta = \frac{4}{5} - \left(-\frac{4}{3}\right) = \frac{12 + 20}{15} = \frac{32}{15}.

Step-by-Step Solution

1
Multiply the numerator and denominator of the given expression by 1+sin⁡θ1 + \sin\theta to simplify it.
cos⁡θ(1+sin⁡θ)(1−sin⁡θ)(1+sin⁡θ)=−3  ⟹  cos⁡θ(1+sin⁡θ)1−sin⁡2θ=−3\frac{\cos\theta(1 + \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)} = -3 \implies \frac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta} = -3
To set up the Pythagorean identity in the denominator.
2
Substitute the Pythagorean identity 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta into the denominator.
cos⁡θ(1+sin⁡θ)cos⁡2θ=1+sin⁡θcos⁡θ=sec⁡θ+tan⁡θ=−3\frac{\cos\theta(1 + \sin\theta)}{\cos^2\theta} = \frac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta = -3
To simplify the expression into basic trigonometric functions.
3
Use the reciprocal identity relationship sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1 to find the difference of secant and tangent.
sec⁡θ−tan⁡θ=1sec⁡θ+tan⁡θ=1−3=−13\sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta} = \frac{1}{-3} = -\frac{1}{3}
To create a system of linear equations for secant and tangent.
4
Solve the system of equations for sec⁡θ\sec\theta and tan⁡θ\tan\theta.
sec⁡θ=−53\sec\theta = -\frac{5}{3} and tan⁡θ=−43\tan\theta = -\frac{4}{3}
Adding the two equations yields 2sec⁡θ=−103  ⟹  sec⁡θ=−532\sec\theta = -\frac{10}{3} \implies \sec\theta = -\frac{5}{3}, and subtracting them yields 2tan⁡θ=−83  ⟹  tan⁡θ=−432\tan\theta = -\frac{8}{3} \implies \tan\theta = -\frac{4}{3}.
5
Find cos⁡θ\cos\theta and sin⁡θ\sin\theta, and calculate the final value of sin⁡θ−tan⁡θ\sin\theta - \tan\theta.
cos⁡θ=−35\cos\theta = -\frac{3}{5}, sin⁡θ=45\sin\theta = \frac{4}{5}, and sin⁡θ−tan⁡θ=45−(−43)=3215\sin\theta - \tan\theta = \frac{4}{5} - \left(-\frac{4}{3}\right) = \frac{32}{15}
Since θ\theta lies in Quadrant II, sine is positive, which is verified by sin⁡θ=tan⁡θ⋅cos⁡θ=(−43)(−35)=45\sin\theta = \tan\theta \cdot \cos\theta = \left(-\frac{4}{3}\right)\left(-\frac{3}{5}\right) = \frac{4}{5}.

Key Concept

Fundamental Trigonometric Identities
Estimated Time:2m 0s
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Fundamental Trigonometric Identities

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