Introduction to Trigonometry

If sin A = 3/5, find cos A and tan A. [CBSE 2025]

Prove that: sin²θ + cos²θ = 1. [CBSE 2024]

Evaluate: (sin 30° + cos 60°) × tan 45°. [CBSE 2023]

If sec θ = 13/5, find other trigonometric ratios. [CBSE 2022]

Express sin 67° + cos 75° in terms of ratios of angles between 0° and 45°. [CBSE 2021]

sin θ = Perpendicular/Hypotenuse\ncos θ = Base/Hypotenuse\ntan θ = Perpendicular/Base\ncot θ = 1/tan θ\nsec θ = 1/cos θ\ncosec θ = 1/sin θ

sin²θ + cos²θ = 1\n1 + tan²θ = sec²θ\n1 + cot²θ = cosec²θ

0°: sin=0, cos=1, tan=0\n30°: sin=1/2, cos=√3/2, tan=1/√3\n45°: sin=1/√2, cos=1/√2, tan=1\n60°: sin=√3/2, cos=1/2, tan=√3\n90°: sin=1, cos=0, tan=∞

sin(90°-θ) = cos θ\ncos(90°-θ) = sin θ\ntan(90°-θ) = cot θ

In right triangle with angle θ: sin θ = p/h, cos θ = b/h. Then sin²θ + cos²θ = p²/h² + b²/h² = (p²+b²)/h². By Pythagoras theorem: p² + b² = h². Therefore sin²θ + cos²θ = h²/h² = 1.

Start with sin²θ + cos²θ = 1. Divide both sides by cos²θ: sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ. This gives: tan²θ + 1 = sec²θ.

In right triangle: sin θ = p/h, cos θ = b/h\nsin²θ + cos²θ = p²/h² + b²/h² = (p²+b²)/h²\nBy Pythagoras: p² + b² = h²\nTherefore: sin²θ + cos²θ = h²/h² = 1

Divide sin²θ + cos²θ = 1 by cos²θ:\nsin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ\ntan²θ + 1 = sec²θ