Question

If \( \sin A = -\frac{60}{61} \), \( \cot B = -\frac{40}{9} \) and neither A nor B is in \( 4^{\text{th}} \) quadrant then \( 6\cot A + 4\sec B = \)

• A. 5
• B. \( \frac{26}{5} \)
• C. -3
• D. 3

Hint:

Memorize common Pythagorean triples like (11, 60, 61) and (9, 40, 41) to save calculation time. Always double-check the sign based on the quadrant.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We need to determine the quadrants for angles A and B to assign the correct signs to the required trigonometric ratios (\( \cot A \) and \( \sec B \)).
Step 2: Key Formula or Approach:

1. ASTC Rule (All, Sin, Tan, Cos positive in Q1, Q2, Q3, Q4 respectively). 2. Pythagorean identities to find missing sides of the reference triangles.
Step 3: Detailed Explanation:

Analyze Angle A: \( \sin A \textless 0 \) implies A is in Q3 or Q4. Since A is not in Q4, A must be in Quadrant III. In Q3, \( \cot A \) is positive. Given \( \sin A = -\frac{60}{61} \) (Opposite/Hypotenuse). Adjacent side \( = \sqrt{61^2 - 60^2} = \sqrt{121} = 11 \). \[ \cot A = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{11}{60} \] Analyze Angle B: \( \cot B \textless 0 \) implies B is in Q2 or Q4. Since B is not in Q4, B must be in Quadrant II. In Q2, \( \sec B \) is negative. Given \( \cot B = -\frac{40}{9} \) (Adjacent/Opposite). Hypotenuse \( = \sqrt{40^2 + 9^2} = \sqrt{1600 + 81} = \sqrt{1681} = 41 \). \[ \sec B = \frac{\text{Hypotenuse}}{\text{Adjacent}} = -\frac{41}{40} \] Calculate Expression: \[ 6\cot A + 4\sec B = 6\left(\frac{11}{60}\right) + 4\left(-\frac{41}{40}\right) \] \[ = \frac{11}{10} - \frac{41}{10} = \frac{-30}{10} = -3 \]
Step 4: Final Answer:

The value is -3.