Trigonometric Ratios Formula 10th Class PDF

3. 𝒄𝒐𝒔𝜽. 𝒔𝒆𝒄𝜽 = 𝟏  Trigonometric Ratios 1 𝑐𝑜𝑠𝜃 = 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐...

3. 𝒄𝒐𝒔𝜽. 𝒔𝒆𝒄𝜽 = 𝟏  Trigonometric Ratios 1 𝑐𝑜𝑠𝜃 = 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑝 𝑠𝑒𝑐𝜃 1. 𝑠𝑖𝑛𝑒 (𝑠𝑖𝑛𝜃) = = 1 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ℎ 𝑠𝑒𝑐𝜃 = 𝑏𝑎𝑠𝑒 𝑏 𝑐𝑜𝑠𝜃 2. 𝑐𝑜𝑠𝑖𝑛𝑒 (𝑐𝑜𝑠𝜃) = = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ℎ  Square relation 3. 𝑡𝑎𝑛𝑔𝑒𝑛𝑡(𝑡𝑎𝑛𝜃) = 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 = 𝑝 1. 𝒔𝒊𝒏𝟐 𝜽 + 𝒄𝒐𝒔𝟐 𝜽 = 𝟏 𝑏𝑎𝑠𝑒 𝑏 𝑠𝑖𝑛2 𝜃 = 1 − 𝑐𝑜𝑠2 𝜃; 𝑠𝑖𝑛𝜃 = 1 − 𝑐𝑜𝑠2 𝜃 𝑏𝑎𝑠𝑒 𝑏 4. 𝑐𝑜𝑡𝑒𝑛𝑔𝑒𝑛𝑡 (𝑐𝑜𝑡𝜃) = = 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑝 cos 2 𝜃 = 1 − sin2 𝜃 ; 𝑐𝑜𝑠𝜃 = 1 − sin2 𝜃 5. 𝑠𝑒𝑐𝑒𝑛𝑡 (𝑠𝑒𝑐𝜃) = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = ℎ 2. 𝒔𝒆𝒄𝟐 𝜽 − 𝒕𝒂𝒏𝟐 𝜽 = 𝟏 𝑏𝑎𝑠𝑒 𝑏 𝑠𝑒𝑐 2 𝜃 = 1 + 𝑡𝑎𝑛2 𝜃; 𝑠𝑒𝑐𝜃 = 1 + 𝑡𝑎𝑛2 𝜃 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ℎ 6. 𝑐𝑜𝑠𝑒𝑐𝑒𝑛𝑡 (𝑐𝑜𝑠𝑒𝑐𝜃) = = 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑝 tan2 𝜃 = sec 2 𝜃 − 1 ; 𝑡𝑎𝑛𝜃 = sec 2 𝜃 − 1 3. 𝒄𝒐𝒔𝒆𝒄𝟐𝜽 − 𝒄𝒐𝒕𝟐 𝜽 = 𝟏  Pythagoras theorem 𝑐𝑜𝑠𝑒𝑐 2 𝜃 = 1 + 𝑐𝑜𝑡 2 𝜃; 𝑐𝑜𝑠𝑒𝑐𝜃 = 1 + 𝑐𝑜𝑡 2 𝜃 1. ℎ2 = 𝑝 2 + 𝑏 2 ; ℎ = 𝑝2 + 𝑏2 𝑐𝑜𝑡 2 𝜃 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 1; 𝑐𝑜𝑡𝜃 = 𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 1 p h  Complementary Angles 2. 𝑝2 = ℎ2 − 𝑏 2 ; 𝑝 = ℎ2 − 𝑏 2 90° 1. 𝑠𝑖𝑛 90° − 𝜃 = 𝑐𝑜𝑠𝜃 3. 𝑏 2 = ℎ2 − 𝑝 2 ; 𝑏 = ℎ2 − 𝑝 2 𝜃 2. 𝑐𝑜𝑠 90° − 𝜃 = 𝑠𝑖𝑛𝜃 b 3. 𝑡𝑎𝑛 90° − 𝜃 = 𝑐𝑜𝑡𝜃  Reciprocal Relations 4. 𝑐𝑜𝑡 90° − 𝜃 = 𝑡𝑎𝑛𝜃 1. 𝒔𝒊𝒏𝜽. 𝒄𝒐𝒔𝒆𝒄𝜽 = 𝟏 5. 𝑠𝑒𝑐 90° − 𝜃 = 𝑐𝑜𝑠𝑒𝑐𝜃 1 𝑠𝑖𝑛𝜃 = 6. 𝑐𝑜𝑠𝑒𝑐 90° − 𝜃 = 𝑠𝑒𝑐𝜃 𝑐𝑜𝑠𝑒𝑐𝜃 1 𝑐𝑜𝑠𝑒𝑐𝜃 =  Quotient relation of T. Ratios 𝑠𝑖𝑛𝜃 2. 𝒕𝒂𝒏𝜽. 𝒄𝒐𝒕𝜽 = 𝟏 𝑠𝑖𝑛 𝜃 1. 𝑡𝑎𝑛𝜃 = 1 𝑐𝑜𝑠 𝜃 𝑡𝑎𝑛𝜃 = 𝑐𝑜𝑠 𝜃 𝑐𝑜𝑡𝜃 2. 𝑐𝑜𝑡𝜃 = 1 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑡𝜃 = 𝑡𝑎𝑛𝜃  Table for T-Ratios of 0°, 30°, 45°, 60°, 90° 𝝅 𝝅 𝝅 𝝅 𝟎° 𝟑𝟎° 𝟒𝟓° 𝟔𝟎° 𝟗𝟎° 𝟔 𝟒 𝟑 𝟐 1 1 3 𝒔𝒊𝒏𝜽 0 2 2 1 2 3 1 1 𝒄𝒐𝒔𝜽 1 2 2 0 2 1 𝒕𝒂𝒏𝜽 0 3 1 3 ∞ 2 𝒄𝒐𝒔𝒆𝒄𝜽 ∞ 2 2 3 1 2 𝒔𝒆𝒄𝜽 1 3 2 2 ∞ 1 𝒄𝒐𝒕𝜽 ∞ 3 1 3 0  Trigonometric Property For 11th Class 𝐴 𝐴 𝐴 SET - I 2. 𝑐𝑜𝑠𝐴 = 𝑐𝑜𝑠 2 − 𝑠𝑖𝑛2 2 = 2𝑐𝑜𝑠 2 2 − 1 = 1 − 2 1. 𝑠𝑖𝑛 𝐴 + 𝐵 = 𝑠𝑖𝑛 𝐴. 𝐶𝑜𝑠𝐵 + 𝐶𝑜𝑠𝐴. 𝑆𝑖𝑛𝐵 𝐴 𝐴 1−𝑡𝑎 𝑛 2 2. 𝑠𝑖𝑛 𝐴 − 𝐵 = 𝑠𝑖𝑛 𝐴. 𝑐𝑜𝑠 𝐵 − 𝑐𝑜𝑠 𝐴. 𝑠𝑖𝑛 𝐵 2 𝑠𝑖𝑛2 2 = 2 𝐴 1+𝑡𝑎 𝑛 2 2 3. 𝑐𝑜𝑠 𝐴 + 𝐵 = 𝑐𝑜𝑠𝐴. 𝑐𝑜𝑠𝐵 − 𝑠𝑖𝑛 𝐴. 𝑠𝑖𝑛𝐵 𝐴 2 𝑡𝑎𝑛 2 4. 𝑐𝑜𝑠 𝐴 − 𝐵 = 𝑐𝑜𝑠 𝐴. 𝑐𝑜𝑠 𝐵 + 𝑠𝑖𝑛 𝐴. 𝑠𝑖𝑛 𝐵 3. 𝑡𝑎𝑛 𝐴 = 𝐴 1−𝑡𝑎 𝑛 2 2 𝑡𝑎𝑛 𝐴+𝑡𝑎𝑛𝐵 5. 𝑡𝑎𝑛 𝐴 + 𝐵 = SET - VI 1−𝑡𝑎𝑛𝐴.𝑡𝑎𝑛𝐵 𝑡𝑎𝑛 𝐴−𝑡𝑎𝑛 𝐵 6. 𝑡𝑎𝑛 𝐴 − 𝐵 = 1. 𝑠𝑖𝑛3𝐴 = 3𝑠𝑖𝑛 𝐴 − 4 𝑠𝑖𝑛3 𝐴 1+𝑡𝑎𝑛𝐴.𝑡𝑎𝑛 𝐵 3 𝑠𝑖𝑛 𝐴 − 𝑠𝑖𝑛 3𝐴 7. 𝑐𝑜𝑡 𝐴 + 𝐵 = 𝑐𝑜𝑡 𝐵.𝑐𝑜𝑡 𝐴−1 ∴ 𝑠𝑖𝑛3 𝐴 = 𝑐𝑜𝑡 𝐵+𝑐𝑜𝑡 𝐴 4 𝑐𝑜𝑡 𝐵.𝑐𝑜𝑡 𝐴+1 2. 𝑐𝑜𝑠 3𝐴 = 4 𝑐𝑜𝑠3 𝐴 − 3 𝑐𝑜𝑠 𝐴 8. 𝑐𝑜𝑡 𝐴 − 𝐵 = 𝑐𝑜𝑡 𝐵−𝑐𝑜𝑡 𝐴 3 𝑐𝑜𝑠 𝐴 + 𝑐𝑜𝑠 3𝐴 9. 𝑡𝑎𝑛 𝐴 + 𝐵 + 𝐶 = ∴ 𝑐𝑜𝑠 3 𝐴 = 4 𝑡𝑎𝑛𝐴 +𝑡𝑎𝑛𝐵 +𝑡𝑎𝑛 𝐶−𝑡𝑎𝑛𝐴.𝑡𝑎𝑛𝐵.𝑡𝑎𝑛𝐶 3 𝑡𝑎𝑛 𝐴−𝑡𝑎 𝑛 3 𝐴 3. 𝑡𝑎𝑛3𝐴 = 1−𝑡𝑎𝑛𝐴.𝑡𝑎𝑛𝐵 −𝑡𝑎𝑛𝐵.𝑡𝑎𝑛𝐶 −𝑡𝑎𝑛𝐶.𝑡𝑎𝑛𝐴 1−3 𝑡𝑎 𝑛 2 𝐴 𝑐𝑜 𝑡 3 𝐴−3𝑐𝑜𝑡 𝐴 SET - II 4. 𝑐𝑜𝑡3𝐴 = 3 𝑐𝑜 𝑡 2 𝐴−1 I. 𝑠𝑖𝑛 𝐴 + 𝐵 + 𝑠𝑖𝑛 𝐴 − 𝐵 = 2𝑠𝑖𝑛𝐴. 𝑐𝑜𝑠𝐵 SET - VII II. 𝑠𝑖𝑛 𝐴 + 𝐵 − 𝑠𝑖𝑛 𝐴 − 𝐵 = 2𝑐𝑜𝑠𝐴. 𝑠𝑖𝑛𝐵 1−𝑐𝑜𝑠 2𝐴 1−𝑐𝑜𝑠 2𝐴 1. 𝑠𝑖𝑛2 𝐴 = ; 𝑠𝑖𝑛𝐴 = III. 𝑐𝑜𝑠 𝐴 + 𝐵 + 𝑐𝑜𝑠 𝐴 − 𝐵 = 2𝑐𝑜𝑠𝐴. 𝑐𝑜𝑠𝐵 2 2 IV. 𝑐𝑜𝑠 𝐴 + 𝐵 − 𝑐𝑜𝑠 𝐴 − 𝐵 = −2𝑠𝑖𝑛𝐴. 𝑠𝑖𝑛𝐵 1+𝑐𝑜𝑠 2 𝐴 1 +𝑐𝑜𝑠 2𝐴 2. 𝑐𝑜𝑠2 𝐴 = ; 𝑐𝑜𝑠 𝐴 = 2 2 V. or 𝑐𝑜𝑠 𝐴 − 𝐵 − 𝑐𝑜𝑠 𝐴 + 𝐵 = 2𝑠𝑖𝑛𝐴. 𝑠𝑖𝑛𝐵 𝐴 1−𝑐𝑜𝑠 𝐴 𝐴 1−𝑐𝑜𝑠 𝐴 3. 𝑠𝑖𝑛2 = ; 𝑠𝑖𝑛 = 2 2 2 2 SET - III 𝐴 1 +𝑐𝑜𝑠 𝐴 𝐴 1+𝑐𝑜𝑠 𝐴 𝐶+𝐷 𝐶−𝐷 4. 𝑐𝑜𝑠2 = ; 𝑐𝑜𝑠 = I. 𝑠𝑖𝑛𝐶 + 𝑠𝑖𝑛𝐷 = 2 𝑠𝑖𝑛. 𝑐𝑜𝑠 2 2 2 2 2 2 𝐶+𝐷 𝐶−𝐷 1−𝑐𝑜𝑠 2𝐴 1−𝑐𝑜𝑠 2𝐴 II. 𝑠𝑖𝑛𝐶 − 𝑠𝑖𝑛𝐷 = 2𝑐𝑜𝑠 2. 𝑠𝑖𝑛 2 5. 𝑡𝑎𝑛2 𝐴 = ; 𝑡𝑎𝑛 𝐴 = 1+𝑐𝑜𝑠 2𝐴 1+𝑐𝑜𝑠 2𝐴 𝐶+𝐷 𝐶−𝐷 III. 𝑐𝑜𝑠 𝐶 + 𝑐𝑜𝑠 𝐷 = 2 𝑐𝑜𝑠. 𝑐𝑜𝑠 2 2 2 6. 1 + 𝑠𝑖𝑛 2𝑥 = 𝑐𝑜𝑠 𝑥 + 𝑠𝑖𝑛 𝑥 𝐶+𝐷 𝐶−𝐷 2 IV. 𝑐𝑜𝑠 𝐶 − 𝑐𝑜𝑠 𝐷 = −2 𝑠𝑖𝑛. 𝑠𝑖𝑛 7. 1 − 𝑠𝑖𝑛 2𝑥 = 𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛 𝑥 2 2 𝑥 𝑥 2 𝐶+𝐷 𝐷−𝐶 8. 1 + 𝑠𝑖𝑛 𝑥 = 𝑐𝑜𝑠 + 𝑠𝑖𝑛 or, 2 𝑠𝑖𝑛. 𝑠𝑖𝑛. 2 2 2 2 𝑥 𝑥 2 SET - IV 9. 1 − 𝑠𝑖𝑛 𝑥 = 𝑐𝑜𝑠 − 𝑠𝑖𝑛 2 2 2 𝑡𝑎𝑛 𝐴 I. 𝑠𝑖𝑛2𝐴 = 2 𝑠𝑖𝑛 𝐴. 𝑐𝑜𝑠 𝐴 = 1+𝑡𝑎𝑛 𝑥 𝜋 1+𝑡𝑎 𝑛 2 𝐴 10. = 𝑡𝑎𝑛 +𝑥 1−𝑡𝑎𝑛 𝑥 4 II. 𝑐𝑜𝑠2𝐴 = 𝑐𝑜𝑠 2 𝐴 − 𝑠𝑖𝑛2 𝐴 = 2𝑐𝑜𝑠2 𝐴 − 1 = 1 − 1−𝑡𝑎𝑛 𝑥 𝜋 2𝑠𝑖𝑛2 𝐴 = 1−𝑡𝑎 𝑛 2 𝐴 11. = 𝑡𝑎𝑛 − 𝑥 1+𝑡𝑎 𝑛 2 𝐴 1+𝑡𝑎𝑛𝑥 4 2 𝑡𝑎𝑛 𝐴 𝑥 III. 𝑡𝑎𝑛 2𝐴 = 1−𝑡𝑎 𝑛 2 𝐴 1+𝑡𝑎𝑛 2 𝜋 𝑥 12. 𝑥 = 𝑡𝑎𝑛 +2 1−𝑡𝑎𝑛 2 4 𝑐𝑜 𝑡 2 𝐴−1 IV. 𝑐𝑜𝑡 2𝐴 = 𝑥 𝑐𝑜𝑡 2𝐴 1−𝑡𝑎𝑛 2 𝜋 𝑥 SET - V 13. 𝑥 = 𝑡𝑎𝑛 −2 1+𝑡𝑎𝑛 2 4 𝐴 𝐴 𝐴 2 𝑡𝑎𝑛 2 1. 𝑠𝑖𝑛𝐴 = 2𝑠𝑖𝑛. 𝑐𝑜𝑠 = 𝐴 2 2 1 +𝑡𝑎 𝑛 2 2