Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Check - Trigonometry Class 11 - All Concepts


Last updated at Feb. 12, 2020 by Teachoo
Learn All Concepts of Chapter 2 Class 11 Relations and Function - FREE. Check - Trigonometry Class 11 - All Concepts
Transcript
Ex 3.3, 3 Prove that cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 = 6 Taking L.H.S. cot2 π/6 + cosec 5π/6 + 3 tan2 π/6 Putting π = 180° = cot2(180/6) + cosec((5 ×180)/6) + 3 tan2(180/6) = cot2 30° + cosec (150°) + 3tan2 30° Here, tan 30° = 1/√3 cot 30° = 1/tan〖30°〗 = 1/(1/√3) = √3 For cosec 150° , First, Finding sin 150° sin 150° = sin (180 – 30° ) = sin 30° = 1/2 cosec 150° = 1/sin〖150°〗 = 1/(1/2) = 2 Putting values = cot2 30° + cosec (150°) + 3tan2 30° = (√3)2 + 2 + 3 × (1/√3)^2 = 3 + 2 + 3 × 1/3 = 3 + 2 + 1 = 6 = R.H.S Hence proved
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Example 8
Ex 3.2, 9 Important
Ex 3.2, 8 Important
Ex 3.2, 10 Important
Example 9 Important
Ex 3.2, 6
Ex 3.2, 7 Important
Example 10
Ex 3.3, 1
Ex 3.3, 2 Important
Ex 3.3, 3 Important You are here
Ex 3.3, 4
Ex 3.3, 8 Important
Ex 3.3, 9 Important
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