Introduction to Trigonometry
Practice Questions

Opposite = 3, Hypotenuse = 5 ⇒ Adjacent = 4. tan θ = 3/4.

sin 30° = 1/2, cos 60° = 1/2. Sum = 1/2 + 1/2 = 1.

9(sec² A - tan² A) = 9(1) = 9.

cot A = 8/15 ⇒ Opp = 15, Adj = 8, Hyp = 17. sin A = 15/17.

Both A and R are true and R is correct explanation.

Both are standard fundamental trigonometric identities, but R does not explain A.

A + B = 60°, A - B = 30° ⇒ 2A = 90° ⇒ A = 45°, B = 15°.

2(1)² + (√3/2)² - (√3/2)² = 2 + 3/4 - 3/4 = 2.

A - B = 30°, A + B = 60° ⇒ A = 45°, B = 15°.

sin(90°-23°) + cos(90°-15°) = cos 23° + sin 15°.

(1+sin A)/cos A × (1-sin A) = (1-sin² A)/cos A = cos² A/cos A = cos A.

LHS = [(1-cos θ)/sin θ]² = (1-cos θ)²/sin² θ = (1-cos θ)²/[(1-cos θ)(1+cos θ)] = (1-cos θ)/(1+cos θ).

LHS = [cos² A + (1+sin A)²] / [cos A(1+sin A)] = [cos² A + 1 + 2sin A + sin² A] / [cos A(1+sin A)] = 2(1+sin A)/[cos A(1+sin A)] = 2 sec A.

Convert to sin θ & cos θ: Numerator becomes sin³ θ - cos³ θ = (sin θ - cos θ)(1 + sin θ cos θ). Result = 1 + sec θ cosec θ.

Squaring both sides: 1 + 2sin θ cos θ = 3 ⇒ sin θ cos θ = 1. tan θ + cot θ = (sin² θ + cos² θ)/(sin θ cos θ) = 1/1 = 1.

tan 30° = 5/BC ⇒ BC = 5√3 cm. sin 30° = 5/AC ⇒ AC = 10 cm.

PR² = PQ² + QR² ⇒ (25-QR)² = 25 + QR² ⇒ 625 - 50QR = 25 ⇒ QR = 12 cm, PR = 13 cm.

LHS = sin² A + cosec² A + 2 + cos² A + sec² A + 2 = (sin² A + cos² A) + 4 + (1+cot² A) + (1+tan² A) = 7 + tan² A + cot² A.

LHS = (cos² A/sin A)(sin² A/cos A) = sin A cos A. RHS = 1 / [1/(sin A cos A)] = sin A cos A.

p² = (1+sin θ)²/cos² θ. p²-1 = 2sin θ(1+sin θ)/cos² θ. p²+1 = 2(1+sin θ)/cos² θ. Ratio = sin θ.