Question

Difficulty: MediumFundamental Trigonometric Identities

For an angle θ\theta satisfying π2<θ<π\frac{\pi}{2} < \theta < \pi, if secθ=135\sec \theta = -\frac{13}{5}, what is the value of 12(cotθ+cscθ)12(\cot \theta + \csc \theta)?

Answer: 8

Answer

8
For an angle θ\theta in Quadrant II (π2<θ<π\frac{\pi}{2} < \theta < \pi), cosine is negative and sine is positive. Given secθ=135\sec \theta = -\frac{13}{5}, the reciprocal identity gives cosθ=513\cos \theta = -\frac{5}{13}. The Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 yields sinθ=1(513)2=1213\sin \theta = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \frac{12}{13}. Then cotθ=cosθsinθ=512\cot \theta = \frac{\cos \theta}{\sin \theta} = -\frac{5}{12} and cscθ=1sinθ=1312\csc \theta = \frac{1}{\sin \theta} = \frac{13}{12}. Adding these values gives cotθ+cscθ=812=23\cot \theta + \csc \theta = \frac{8}{12} = \frac{2}{3}. Multiplying by 12 yields the final value of 8.

Step-by-Step Solution

1
Find cosθ\cos \theta from secθ\sec \theta
cosθ=513\cos \theta = -\frac{5}{13}
By definition of the reciprocal trigonometric identity, cosθ=1secθ\cos \theta = \frac{1}{\sec \theta}.
2
Calculate sinθ\sin \theta using the Pythagorean identity
sinθ=1213\sin \theta = \frac{12}{13}
In Quadrant II (π2<θ<π\frac{\pi}{2} < \theta < \pi), sine is positive. Applying sinθ=1cos2θ\sin \theta = \sqrt{1 - \cos^2 \theta} gives 125169=1213\sqrt{1 - \frac{25}{169}} = \frac{12}{13}.
3
Find cotθ\cot \theta and cscθ\csc \theta
\cot \theta = -\frac{5}{12} \text{ and } \csc \theta = \frac{13}{12}
Using quotient identity cotθ=cosθsinθ\cot \theta = \frac{\cos \theta}{\sin \theta} and reciprocal identity cscθ=1sinθ\csc \theta = \frac{1}{\sin \theta}.
4
Substitute into the given expression 12(cotθ+cscθ)12(\cot \theta + \csc \theta) and simplify
8
12(512+1312)=12(812)=812\left(-\frac{5}{12} + \frac{13}{12}\right) = 12\left(\frac{8}{12}\right) = 8.

Key Concept

Pythagorean, quotient, and reciprocal identities with quadrant sign analysis
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