If 4tanθ − 3 = 0, then the value of (1 − cos2θ) ⁄ (1 + cos2θ) is:

Correct — A. From 4tanθ − 3 = 0, tanθ = 3⁄4.

Use the two double-angle forms of cos2θ:
1 − cos2θ = 1 − (1 − 2sin²θ) = 2sin²θ
1 + cos2θ = 1 + (2cos²θ − 1) = 2cos²θ
So the ratio = 2sin²θ ⁄ 2cos²θ = tan²θ

tan²θ = (3⁄4)² = 9⁄16 → option (a).

Why the others are wrong

• (b)
  4⁄3 is cotθ, the reciprocal of tanθ. The expression reduces to a square, so a first-power value cannot be it.
• (c)
  7⁄15 is neither tan²θ nor cos2θ: with tanθ = 3⁄4, cos2θ = (1 − 9⁄16)⁄(1 + 9⁄16) = 7⁄25, and the required ratio is 9⁄16.
• (d)
  The ratio equals 1 only when tan²θ = 1, that is θ = 45°. Here tanθ = 3⁄4 is below 1, so its square is below 1 too.

Two identities collapse the whole expression. cos2θ = 1 − 2sin²θ turns the numerator into 2sin²θ, and cos2θ = 2cos²θ − 1 turns the denominator into 2cos²θ.

Dividing cancels the 2 and leaves tan²θ. Because the expression reduces to a function of tanθ alone, the angle itself is never needed — and since the result is a square, the sign of tanθ cannot matter.

Key facts

• cos2θ has three standard forms: 1 − 2sin²θ, 2cos²θ − 1, and cos²θ − sin²θ.
• (1 − cos2θ)⁄(1 + cos2θ) = tan²θ, and the reciprocal expression equals cot²θ.
• In terms of tanθ: cos2θ = (1 − tan²θ)⁄(1 + tan²θ) and sin2θ = 2tanθ⁄(1 + tan²θ).
• tanθ = 3⁄4 is the 3-4-5 right triangle, so an acute θ has sinθ = 3⁄5 and cosθ = 4⁄5.

Common traps

• Solving 4tanθ − 3 = 0 for the angle, which the expression never requires.
• Reporting tanθ = 3⁄4 when the identity delivers the square.
• Confusing this with the half-angle form (1 − cosθ)⁄(1 + cosθ), which equals tan²(θ⁄2).

No directly related past PYQ was found.