1. Class 11
2. Important Question for exams Class 11

Transcript

Ex 3.3, 5 Find the value of: sin 75° sin 75° = sin (45° + 30°) We know that sin (x + y) = sin x cos y + cos x sin y Hence putting x = 45° and y = 30° So, sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° sin 75° = 1/√2 × √3/2 + 1/√2 × 1/2 = 1/√2 ( √3/2 + 1/2 ) = (√3 + 1)/(2√2) Ex 3.3, 5 Find the value of: (ii) tan 15° tan 15° = tan (45° – 30°) Using tan (x - y) = (𝑡𝑎𝑛𝑥 − 𝑡𝑎𝑛⁡𝑦)/(1+𝑡𝑎𝑛 𝑥 𝑡𝑎𝑛⁡𝑦 ) Here putting x = 45° and y = 30° Putting value tan (45° – 30° ) = (tan 45"°" − 〖 tan〗⁡30"°" )/(1 + tan 45"°" tan⁡30"°" ) tan 15° = (1 − 1/√3)/(1 + 1 × 1/√3) = ((√3 − 1" " )/√3)/((√3 + 1" " )/√3) = (√3 −1)/√3 × √3/(√3 + 1) = (√3 − 1)/(√3 + 1) Rationalizing = (√3 − 1)/(√3 + 1) × (√3 − 1)/(√3 − 1) = (√3 − 1)2/(√3 + 1)(√3 − 1) Using (a – b )2 = a2 + b2 – 2ab = ((√3)2 +(1)2 −2" " × √3 × 1)/(√3 + 1)(√3 − 1) = (3 + 1 − 2√3)/(√(3 )+ 1)(√3 − 1) Using (a – b ) (a + b) = a2 – b2 = (4 − 2√3)/((√3)2 − (1)2) = (4 − 2√3)/(3 − 1) = (4 − 2√3)/2 = (2 (2 − √(3 )))/2 = 2 – √3 Hence tan 15° = 2 – √3

Class 11
Important Question for exams Class 11