If tan θ = 5⁄8, then find the value of [(1 + cos θ)(1 − cos θ)] ⁄ [(1 + sin θ)(1 − sin θ)].

Correct — D. Both brackets are a difference of squares, so expand them before the value of tan θ goes anywhere near the page.

(1 + cos θ)(1 − cos θ) = 1 − cos²θ = sin²θ
(1 + sin θ)(1 − sin θ) = 1 − sin²θ = cos²θ

sin²θ ⁄ cos²θ = tan²θ
= (5⁄8)² = 25⁄64 → option (d)

Why the others are wrong

• (a)
  5⁄64 squares only the denominator of 5⁄8. Squaring a fraction squares both parts, which gives 25⁄64.
• (b)
  2⁄25 is neither tan²θ (25⁄64) nor cot²θ (64⁄25) — it only borrows the 25, and no step of this simplification produces it.
• (c)
  64⁄25 is cot²θ, the ratio read upside down. The numerator collapses to sin²θ and the denominator to cos²θ, so the quotient is tan²θ, not its reciprocal.

Two identities do all the work. 1 − cos²θ = sin²θ and 1 − sin²θ = cos²θ are the Pythagorean identity rearranged, and any (1 + x)(1 − x) bracket in a trigonometric expression is an invitation to use one of them.

Once the expression reduces to sin²θ ⁄ cos²θ, the given ratio drops straight in: tan θ = 5⁄8 makes tan²θ = 25⁄64.

You never need sin θ and cos θ separately. Build the 5–8–√89 right triangle if you like and you get sin θ = 5⁄√89, cos θ = 8⁄√89 and the same 25⁄64 — three extra lines for one answer.

Key facts

• (1 + cos θ)(1 − cos θ) = 1 − cos²θ = sin²θ.
• (1 + sin θ)(1 − sin θ) = 1 − sin²θ = cos²θ.
• The whole expression therefore equals tan²θ for every θ with cos θ ≠ 0.
• tan θ = 5⁄8 corresponds to a right triangle with legs 5 and 8 and hypotenuse √89.

Common traps

• Inverting the fraction and answering cot²θ = 64⁄25.
• Squaring only the numerator or only the denominator of 5⁄8.
• Computing sin θ and cos θ separately when the expression never needs either.

No directly related past PYQ was found.