Simplify: (tan² θ−sin² θ)⁄(2+tan² θ+cot² θ)

Correct — B. Simplify the top and the bottom separately, then divide.

Numerator, take out sin²θ: tan²θ − sin²θ = sin²θ(sec²θ − 1)
Use sec²θ − 1 = tan²θ: = sin²θ · tan²θ = sin⁴θ⁄cos²θ

Denominator, split 2 as 1 + 1: (1 + tan²θ) + (1 + cot²θ) = sec²θ + cosec²θ
Add the fractions: 1⁄cos²θ + 1⁄sin²θ = (sin²θ + cos²θ)⁄(sin²θ cos²θ) = 1⁄(sin²θ cos²θ)

Divide: sin⁴θ⁄cos²θ × sin²θ cos²θ = sin⁶θ → option (b)

Why the others are wrong

• (a)
  sec⁶ θ — sec⁶θ is at least 1 for every θ, yet the expression is small: at θ = 45° it equals (1 − 1⁄2) ⁄ (2 + 1 + 1) = 1⁄8, while sec⁶45° = (√2)⁶ = 8.
• (c)
  cos² θ — cos²θ leaves a cosine that the working cancels completely. After the division only sines remain, and at θ = 45°, cos²θ = 1⁄2 while the expression is 1⁄8.
• (d)
  sin² θ — sin²θ stops too early. The numerator is already sin⁴θ⁄cos²θ, and dividing by the denominator multiplies by sin²θ cos²θ, taking the power of sine to 6. At θ = 45°, sin²θ = 1⁄2, not 1⁄8.

Two Pythagorean identities do the work: sec²θ − 1 = tan²θ and 1 + cot²θ = cosec²θ.

A lone 2 next to tan²θ + cot²θ can be split as 1 + 1, one for each square. The denominator then becomes sec²θ + cosec²θ, which equals sec²θ · cosec²θ and also (tanθ + cotθ)².

Writing everything in sin and cos and cancelling is slower, but it works on any expression of this kind.

Key facts

• sec²θ − 1 = tan²θ and cosec²θ − 1 = cot²θ.
• 2 + tan²θ + cot²θ = (tanθ + cotθ)² = sec²θ · cosec²θ.
• tan²θ − sin²θ = tan²θ · sin²θ.

Common traps

• Treating tan²θ + cot²θ as 1: it is tanθ · cotθ that equals 1, not the sum of the squares.
• Stopping once the numerator is simplified: dividing by the denominator still multiplies by sin²θ cos²θ.

No directly related past PYQ was found.