If tan θ = 8⁄15, and θ is an acute angle, then the value of √(1 − sin θ) ⁄ √(1 + sin θ) is:

Correct — D. The stem is an image: given tan θ = 8⁄15 with θ acute, find √(1 − sin θ) ⁄ √(1 + sin θ), which is the same as √((1 − sin θ)⁄(1 + sin θ)).

tan θ = 8⁄15 fixes a right triangle with legs 8 and 15.
Hypotenuse = √(8² + 15²) = √289 = 17
sin θ = 8⁄17

Substitute.
1 − sin θ = 9⁄17
1 + sin θ = 25⁄17
Their ratio = (9⁄17) ÷ (25⁄17) = 9⁄25

√(9⁄25) = 3⁄5, and 3⁄5 is what option (d) prints → option (d)

Identity route: for an acute angle, √((1 − sin θ)⁄(1 + sin θ)) = sec θ − tan θ = 17⁄15 − 8⁄15 = 3⁄5.

Why the others are wrong

• (a)
  Option (a) prints 4⁄5. Worked backwards that value needs sin θ = 9⁄41, not the 8⁄17 that tan θ = 8⁄15 forces.
• (b)
  Option (b) prints 2⁄5. Backwards it needs sin θ = 21⁄29, an angle whose tangent is 21⁄20 — nothing like the 8⁄15 given.
• (c)
  Option (c) prints 1⁄5, which needs sin θ = 12⁄13. That is the 5-12-13 angle, with tan θ = 12⁄5, far steeper than this one.

A single trigonometric ratio for an acute angle fixes the whole right triangle, and with it every other ratio.

tan θ = 8⁄15 means the opposite side is 8k and the adjacent 15k, so the hypotenuse is 17k and k cancels out of every ratio. Recognising 8-15-17 as a Pythagorean triple saves the surd work.

The expression has a standard collapse. Multiply top and bottom inside the root by (1 − sin θ):

(1 − sin θ)² ⁄ (1 − sin²θ) = (1 − sin θ)² ⁄ cos²θ

The root of that is (1 − sin θ)⁄cos θ = sec θ − tan θ, which stays positive right across the acute range.

Key facts

• 8, 15, 17 is a Pythagorean triple, so tan θ = 8⁄15 gives sin θ = 8⁄17 and cos θ = 15⁄17.
• For an acute angle, √((1 − sin θ)⁄(1 + sin θ)) = sec θ − tan θ.
• Here sec θ − tan θ = 17⁄15 − 8⁄15 = 3⁄5, and sec θ + tan θ = 25⁄15 = 5⁄3.
• Since sec²θ − tan²θ = 1, those two are reciprocals: 3⁄5 × 5⁄3 = 1.

Common traps

• Forgetting the square root and stopping at 9⁄25.
• Reading tan θ = 8⁄15 as sin θ = 8⁄15, which wrongly puts 15 on the hypotenuse.
• Assuming a 3-4-5 or 5-12-13 triangle out of habit.

No directly related past PYQ was found.