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 are positive in Quadrants I, II, III, and IV respectively.
2. Pythagorean identity is used to find the missing side of the reference triangle.

Step 3: Detailed Explanation:

Analyze Angle A:

sin A < 0 implies A lies in Quadrant III or Quadrant IV.
Since A is not in Quadrant IV, A must be in Quadrant III.

In Quadrant III, cot A is positive.

Given:
sin A = -60/61

So, taking opposite = 60 and hypotenuse = 61,
the adjacent side is:
√(612 - 602) = √(3721 - 3600) = √121 = 11

Therefore,
cot A = adjacent / opposite = 11/60

Analyze Angle B:

cot B < 0 implies B lies in Quadrant II or Quadrant IV.
Since B is not in Quadrant IV, B must be in Quadrant II.

In Quadrant II, sec B is negative.

Given:
cot B = -40/9

So, taking adjacent = 40 and opposite = 9,
the hypotenuse is:
√(402 + 92) = √(1600 + 81) = √1681 = 41

Therefore,
sec B = hypotenuse / adjacent = -41/40

Calculate the Expression:

6cot A + 4sec B = 6(11/60) + 4(-41/40)

= 11/10 - 41/10

= -30/10

= -3

Step 4: Final Answer:

The value is -3.