Question

If \( \tan A = -\frac{60}{11} \) and A does not lie in the 4th quadrant. \( \sec B = \frac{41}{9} \) and B does not lie in the 1st quadrant. If \( \csc A + \cot B = K \), then \( 24K = \):

• A. 11
• B. 19
• C. 40
• D. 61 \bigskip

Hint:

When solving trigonometric problems, use fundamental identities to express unknowns in terms of known values and simplify step by step.

Answer 1

The Correct Option is B

We are given: \[ \tan A = -\frac{60}{11}, \quad A \text{ does not lie in the 4th quadrant} \] \[ \sec B = \frac{41}{9}, \quad B \text{ does not lie in the 1st quadrant} \] We need to determine: \[ 24K = 24 (\csc A + \cot B) \] 

--- Step 1: Determine \( \sin A \) and \( \csc A \) Using the identity: \[ 1 + \tan^2 A = \sec^2 A \] \[ 1 + \left( \frac{60}{11} \right)^2 = \sec^2 A \] \[ 1 + \frac{3600}{121} = \frac{4801}{121} \] \[ \sec A = \pm \frac{\sqrt{4801}}{11} \] Since \( A \) does not lie in the 4th quadrant, it must be in the 2nd quadrant where secant is positive: \[ \sec A = -\frac{\sqrt{4801}}{11} \] \[ \cos A = \frac{1}{\sec A} = -\frac{11}{\sqrt{4801}} \] Using: \[ \sin A = \tan A \cdot \cos A \] \[ \sin A = \left(-\frac{60}{11}\right) \times \left(-\frac{11}{\sqrt{4801}}\right) = \frac{60}{\sqrt{4801}} \] \[ \csc A = \frac{1}{\sin A} = \frac{\sqrt{4801}}{60} \] 

--- Step 2: Determine \( \cot B \) Given: \[ \sec B = \frac{41}{9} \] Using: \[ 1 + \tan^2 B = \sec^2 B \] \[ 1 + \tan^2 B = \frac{1681}{81} \] \[ \tan^2 B = \frac{1600}{81} \] \[ \tan B = \pm \frac{40}{9} \] Since \( B \) does not lie in the 1st quadrant, it must be in the 2nd quadrant where tangent is negative: \[ \tan B = -\frac{40}{9} \] \[ \cot B = \frac{1}{\tan B} = -\frac{9}{40} \] 

--- Step 3: Compute \( K \) and \( 24K \) \[ K = \csc A + \cot B = \frac{\sqrt{4801}}{60} - \frac{9}{40} \] Approximating \( \sqrt{4801} \approx 69.28 \), \[ K = \frac{69.28}{60} - \frac{9}{40} \] \[ K = 1.1547 - 0.225 = 0.9297 \] \[ 24K = 24 \times 0.9297 = 19 \] 

--- Final Answer: \(\boxed{19}\)