Question

If \( sin~A=\frac{3}{5} \) and A lies in the second quadrant, then \( \frac{tan~A-sec~A}{cot~A+cosec~A}= \)

• A. \( \frac{2}{5} \)
• B. \( \frac{11}{15} \)
• C. \( \frac{4}{3} \)
• D. \( \frac{3}{2} \)

Hint:

Always double check ASTC signs before performing substitutions:

• Quadrant I: All positive.

• Quadrant II: Sine and Cosecant positive.

• Quadrant III: Tangent and Cotangent positive.

• Quadrant IV: Cosine and Secant positive.

Answer 1

The Correct Option is D

Step 1: Note the quadrant signs.
In the second quadrant only sine and cosecant are positive; cosine, secant, tangent, cotangent are negative.
Step 2: Build the basic ratios.
Given $\sin A = \tfrac{3}{5}$, the third side is $\sqrt{25-9} = 4$. So $\cos A = -\tfrac{4}{5}$.
Step 3: Write all needed ratios.
Then $\tan A = -\tfrac{3}{4}$, $\sec A = -\tfrac{5}{4}$, $\cot A = -\tfrac{4}{3}$, $\csc A = \tfrac{5}{3}$.
Step 4: Work out the top.
$\tan A - \sec A = -\tfrac{3}{4} + \tfrac{5}{4} = \tfrac{1}{2}$.
Step 5: Work out the bottom.
$\cot A + \csc A = -\tfrac{4}{3} + \tfrac{5}{3} = \tfrac{1}{3}$.
Step 6: Divide.
\[ \frac{1/2}{1/3} = \frac{3}{2} \] \[ \boxed{\tfrac{3}{2}} \]