If 2 tan θ = 3, then (3 sin θ − 2 cos θ) ⁄ (3 sin θ + 2 cos θ) is equal to:

Correct — B. The stem is an image: 2 tan θ = 3, so tan θ = 3⁄2.

Divide every term of the fraction by cos θ — legal, because cos θ is not zero when tan θ is finite:
(3 sin θ − 2 cos θ) ⁄ (3 sin θ + 2 cos θ) = (3 tan θ − 2) ⁄ (3 tan θ + 2)

Now substitute tan θ = 3⁄2:
numerator = 3 × 3⁄2 − 2 = 9⁄2 − 4⁄2 = 5⁄2
denominator = 3 × 3⁄2 + 2 = 9⁄2 + 4⁄2 = 13⁄2

The halves cancel:
(5⁄2) ÷ (13⁄2) = 5⁄13 → option (b), the picture showing 5 over 13.

Why the others are wrong

• (a)
  6⁄13 keeps the denominator right but slips in the numerator. 3 tan θ − 2 = 9⁄2 − 4⁄2 = 5⁄2; a numerator of 6⁄2 would mean subtracting 3⁄2 where the stem subtracts 2 cos θ.
• (c)
  6⁄7 cannot come out of this substitution. With tan θ = 3⁄2 the denominator 3 tan θ + 2 is 13⁄2, so 13 is fixed below the line before you ever touch the numerator.
• (d)
  5⁄7 has the numerator right — 5⁄2 — and the denominator wrong. Adding gives 9⁄2 + 4⁄2 = 13⁄2, not 7⁄2, so the correct ratio is 5⁄13.

Numerator and denominator here are both homogeneous of degree one in sin θ and cos θ — every term is a constant times one of them.

For such a fraction you never need θ itself. Divide top and bottom by cos θ and the whole thing collapses into tan θ, which the stem hands you.

The trick runs the other way too: divide by sin θ and it collapses into cot θ. Either works, because the two degrees match and the dividing factor cancels completely.

Key facts

• Dividing a fraction that is homogeneous in sin θ and cos θ by cos θ turns it into an expression in tan θ alone.
• 2 tan θ = 3 gives tan θ = 3⁄2, so 3 tan θ = 9⁄2.
• (3 tan θ − 2) ⁄ (3 tan θ + 2) at tan θ = 3⁄2 equals (5⁄2) ⁄ (13⁄2) = 5⁄13.

Common traps

• Reading the stem as tan θ = 3 rather than 2 tan θ = 3, which lands on 7⁄11 and no option at all.
• Dividing by sin θ and then using cot θ = 3⁄2 instead of 2⁄3.
• Cancelling the halves too early and losing the 2 in the constant terms.

No directly related past PYQ was found.