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Difficulty: HardTrigonometric Ratios and Identities

For an acute angle θ\theta, the expression sin⁡(θ)+cos⁡(θ)sec⁡(θ)+csc⁡(θ)\frac{\sin(\theta) + \cos(\theta)}{\sec(\theta) + \csc(\theta)} is equal to kk. If tan⁡(θ)=3\tan(\theta) = 3, what is the value of kk?

  1. A
    35\frac{3}{5}
  2. 310\frac{3}{10}Answer
  3. C
    33
  4. D
    85\frac{8}{5}

Answer

The correct answer is three-tenths, or 3/10.
The correct answer is 310\frac{3}{10}. By rewriting the denominator using reciprocal identities, we have sec⁡(θ)+csc⁡(θ)=1cos⁡(θ)+1sin⁡(θ)=sin⁡(θ)+cos⁡(θ)sin⁡(θ)cos⁡(θ)\sec(\theta) + \csc(\theta) = \frac{1}{\cos(\theta)} + \frac{1}{\sin(\theta)} = \frac{\sin(\theta) + \cos(\theta)}{\sin(\theta)\cos(\theta)}. Dividing the numerator by this expression simplifies the entire fraction to sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta). Using the reference right triangle for tan⁡(θ)=3\tan(\theta) = 3, we find that sin⁡(θ)=310\sin(\theta) = \frac{3}{\sqrt{10}} and cos⁡(θ)=110\cos(\theta) = \frac{1}{\sqrt{10}}, which gives a product of 310\frac{3}{10}.

Step-by-Step Solution

1
Rewrite the terms in the denominator using their reciprocal definitions.
sec⁡(θ)=1cos⁡(θ)\sec(\theta) = \frac{1}{\cos(\theta)} and csc⁡(θ)=1sin⁡(θ)\csc(\theta) = \frac{1}{\sin(\theta)}, so the expression becomes sin⁡(θ)+cos⁡(θ)1cos⁡(θ)+1sin⁡(θ)\frac{\sin(\theta) + \cos(\theta)}{\frac{1}{\cos(\theta)} + \frac{1}{\sin(\theta)}}.
Expressing secant and cosecant in terms of sine and cosine allows for the simplification of the complex fraction.
2
Combine the terms in the denominator by finding a common denominator.
1cos⁡(θ)+1sin⁡(θ)=sin⁡(θ)+cos⁡(θ)sin⁡(θ)cos⁡(θ)\frac{1}{\cos(\theta)} + \frac{1}{\sin(\theta)} = \frac{\sin(\theta) + \cos(\theta)}{\sin(\theta)\cos(\theta)}.
Finding a common denominator simplifies the addition of the fractional terms.
3
Divide the numerator by the simplified denominator.
sin⁡(θ)+cos⁡(θ)sin⁡(θ)+cos⁡(θ)sin⁡(θ)cos⁡(θ)=sin⁡(θ)cos⁡(θ)\frac{\sin(\theta) + \cos(\theta)}{\frac{\sin(\theta) + \cos(\theta)}{\sin(\theta)\cos(\theta)}} = \sin(\theta)\cos(\theta).
Since θ\theta is an acute angle, the sum sin⁡(θ)+cos⁡(θ)\sin(\theta) + \cos(\theta) is non-zero, allowing it to be canceled from both the numerator and the denominator.
4
Find the values of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) using the given relation tan⁡(θ)=3\tan(\theta) = 3.
Construct a right triangle where the opposite side to angle θ\theta is 33 and the adjacent side is 11. The hypotenuse is 32+12=10\sqrt{3^2 + 1^2} = \sqrt{10}. Therefore, sin⁡(θ)=310\sin(\theta) = \frac{3}{\sqrt{10}} and cos⁡(θ)=110\cos(\theta) = \frac{1}{\sqrt{10}}.
Using a reference right triangle is a direct way to find the values of other trigonometric functions from a known tangent value.
5
Multiply the values of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) to find the value of kk.
k=sin⁡(θ)cos⁡(θ)=(310)(110)=310k = \sin(\theta)\cos(\theta) = \left(\frac{3}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{10}}\right) = \frac{3}{10}.
This yields the final numerical value of the expression.

Key Concept

Simplification of trigonometric expressions using reciprocal identities and finding trigonometric ratios from a given tangent ratio.
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Topics

Trigonometric Ratios and Identities

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