Examkin

Question

Difficulty: Very hardFundamental Trigonometric Identities

For an angle θ\theta in the interval 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi, the expression sec⁡θ−tan⁡θ\sec\theta - \tan\theta is equal to 33. What is the value of csc⁡θ+cot⁡θ\csc\theta + \cot\theta?

Answer: -2

Answer

The value of csc⁡θ+cot⁡θ\csc\theta + \cot\theta is −2-2.
By using the difference of squares on the identity sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, we get (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1. Substituting sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3 gives sec⁡θ+tan⁡θ=13\sec\theta + \tan\theta = \frac{1}{3}. Solving the system of equations gives sec⁡θ=53\sec\theta = \frac{5}{3} and tan⁡θ=−43\tan\theta = -\frac{4}{3}. Since θ\theta lies in Quadrant IV, cos⁡θ=35\cos\theta = \frac{3}{5} and sin⁡θ=−45\sin\theta = -\frac{4}{5}. We then find csc⁡θ=−54\csc\theta = -\frac{5}{4} and cot⁡θ=−34\cot\theta = -\frac{3}{4}, which sum to −2-2.

Step-by-Step Solution

1
Use the Pythagorean identity sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, which factors into (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1.
Since sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3, we have 3(sec⁡θ+tan⁡θ)=1  ⟹  sec⁡θ+tan⁡θ=133(\sec\theta + \tan\theta) = 1 \implies \sec\theta + \tan\theta = \frac{1}{3}.
To establish a second linear equation in terms of sec⁡θ\sec\theta and tan⁡θ\tan\theta.
2
Add and subtract the two equations: sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3 and sec⁡θ+tan⁡θ=13\sec\theta + \tan\theta = \frac{1}{3}.
Adding them gives 2sec⁡θ=103  ⟹  sec⁡θ=532\sec\theta = \frac{10}{3} \implies \sec\theta = \frac{5}{3}. Subtracting the first from the second gives 2tan⁡θ=−83  ⟹  tan⁡θ=−432\tan\theta = -\frac{8}{3} \implies \tan\theta = -\frac{4}{3}.
To isolate the values of sec⁡θ\sec\theta and tan⁡θ\tan\theta.
3
Find cos⁡θ\cos\theta and sin⁡θ\sin\theta using cos⁡θ=1sec⁡θ\cos\theta = \frac{1}{\sec\theta} and sin⁡θ=tan⁡θcos⁡θ\sin\theta = \tan\theta\cos\theta.
cos⁡θ=35\cos\theta = \frac{3}{5} and sin⁡θ=−45\sin\theta = -\frac{4}{5}. Since 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi (Quadrant IV), cosine is positive and sine is negative, which matches these values.
To find the primary trigonometric values needed for the reciprocal functions.
4
Calculate csc⁡θ\csc\theta and cot⁡θ\cot\theta using reciprocal identities.
csc⁡θ=1sin⁡θ=−54\csc\theta = \frac{1}{\sin\theta} = -\frac{5}{4} and cot⁡θ=1tan⁡θ=−34\cot\theta = \frac{1}{\tan\theta} = -\frac{3}{4}.
To obtain the terms of the required sum.
5
Sum the values of csc⁡θ\csc\theta and cot⁡θ\cot\theta.
csc⁡θ+cot⁡θ=−54+(−34)=−84=−2\csc\theta + \cot\theta = -\frac{5}{4} + \left(-\frac{3}{4}\right) = -\frac{8}{4} = -2.
To obtain the final value requested by the question.

Key Concept

Pythagorean and Reciprocal Trigonometric Identities
Rate this question

Topics

Fundamental Trigonometric Identities

More questions from this topic