There is some mistake in this Video. Please check images above.
Ex 3.3, 2 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 13, 2024 by Teachoo
Last updated at Dec. 13, 2024 by Teachoo
There is some mistake in this Video. Please check images above.
Ex 3.3, 2 Prove that 2sin2 π/6 + cosec2 7π/6 cos2 π/3 = 3/2 Solving L.H.S 2sin2 π/6 + cosec2 7π/6 cos2 π/3 Putting π = 180° = 2 sin2 180/6 + cosec2 (7 ×180)/6 cos2 180/3 = 2sin2 30° + cosec2 210° cos2 60° = 2(sin 30°)2 + (cosec 210°)2 (cos 60°)2 For cosec 210° , Lets first calculate sin 210° sin 210° = sin (180° + 30°) = −sin 30° = (−1)/2 So, cosec 210° = 1/sin〖210°〗 = 1/((−1)/2) = 2/(−1 ) = –2 Putting values in equation 2(sin 30°)2 + (cosec 210°)2 (cos 60°)2 = 2 (1/2)^2 + (–2)2 × (1/2)^2 = 2 × 1/4 + 4 × 1/4 = 𝟏/𝟐 + 1 = 1/2 + 1 = 3/2 = R.H.S Hence proved
Ex 3.3
Ex 3.3, 2 Important You are here
Ex 3.3, 3 Important
Ex 3.3, 4
Ex 3.3, 5 (i) Important
Ex 3.3, 5 (ii)
Ex 3.3, 6 Important
Ex 3.3, 7
Ex 3.3, 8 Important
Ex 3.3, 9 Important
Ex 3.3, 10
Ex 3.3, 11 Important
Ex 3.3, 12
Ex 3.3, 13 Important
Ex 3.3, 14
Ex 3.3, 15
Ex 3.3, 16 Important
Ex 3.3, 17
Ex 3.3, 18 Important
Ex 3.3, 19
Ex 3.3, 20
Ex 3.3, 21 Important
Ex 3.3, 22 Important
Ex 3.3, 23 Important
Ex 3.3, 24
Ex 3.3, 25
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo