The statement ${ }^{\prime \prime} \tan _\theta=-\frac{12}{5}, \csc \theta=-\frac{13}{5}$, and the terminal point determined by $\theta$ is in quadrant $3^{\prime \prime}$:
A. cannot be true because $\tan \theta$ must be less than 1.
B. cannot be true because if $\tan \theta=-\frac{12}{5}$, then $\csc \theta= \pm \frac{13}{12}$.
C. cannot be true because $\tan \theta$ is greater than zero in quadrant 3.
D. cannot be true because $12^2+5^2 \neq 1$.
Answer (1)
tan ( θ ) is negative, implying θ is in quadrant 2 or 4.
In quadrant 3, tan ( θ ) is positive, contradicting the given tan ( θ ) = − 5 12 .
Therefore, the statement cannot be true because tan ( θ ) is greater than zero in quadrant 3.
The correct option is C: cannot be true because tan θ is greater than zero in quadrant 3.
C
Explanation
Analyze the given information. We are given that tan ( θ ) = − 5 12 and csc ( θ ) = − 5 13 , and that the terminal point determined by θ is in quadrant 3. We need to determine why this statement cannot be true.
Analyze the sign of tan in quadrant 3. Recall that in quadrant 3, both the x and y coordinates are negative. Therefore, tan ( θ ) = x y must be positive in quadrant 3 since a negative divided by a negative is positive. However, we are given that tan ( θ ) = − 5 12 , which is negative. This creates a contradiction.
Conclude the correct option. Since tan ( θ ) is negative, θ cannot be in quadrant 3. Therefore, the statement cannot be true because tan ( θ ) is greater than zero in quadrant 3.
Analyze the sign of csc in quadrant 3. Alternatively, we can analyze the sign of csc ( θ ) . Recall that csc ( θ ) = s i n ( θ ) 1 = y r , where r is the radius and is always positive. Since csc ( θ ) = − 5 13 < 0 , we must have y < 0 . In quadrant 3, y < 0 , so this condition is satisfied. However, since tan ( θ ) = x y < 0 , we must have 0"> x > 0 , which contradicts the fact that x < 0 in quadrant 3.
Analyze using trigonometric identities. Another approach is to consider the identity 1 + tan 2 ( θ ) = sec 2 ( θ ) . If tan ( θ ) = − 5 12 , then tan 2 ( θ ) = 25 144 . Thus, sec 2 ( θ ) = 1 + 25 144 = 25 169 , so sec ( θ ) = ± 5 13 . Since sec ( θ ) = c o s ( θ ) 1 = x r , we have cos ( θ ) = ± 13 5 . If cos ( θ ) = 13 5 , then 0"> x > 0 . If cos ( θ ) = − 13 5 , then x < 0 . Also, cot ( θ ) = t a n ( θ ) 1 = − 12 5 . Since csc 2 ( θ ) = 1 + cot 2 ( θ ) , we have csc 2 ( θ ) = 1 + 144 25 = 144 169 , so csc ( θ ) = ± 12 13 . This contradicts the given csc ( θ ) = − 5 13 .
Final Answer. Therefore, the correct answer is C.
Examples
Understanding the signs of trigonometric functions in different quadrants is crucial in navigation. For example, if a ship's radar indicates an object at a certain bearing (angle) and the trigonometric functions of that bearing have specific signs, navigators can determine the object's relative position (quadrant) to avoid collisions or to chart a course effectively. This ensures safe and accurate navigation by applying trigonometric principles to real-world scenarios.
