Ex 3.3, 4 - Chapter 3 Class 11 Trigonometric Functions

Last updated at April 16, 2024 by

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Ex 3.3, 4 Prove that 2sin2 3π/4 + 2cos2 π/4 + 2sec2 π/3 = 10 Solving L.H.S 2sin2 3π/4 + 2cos2 π/4 + 2sec2 π/3 Putting π = 180° 2 sin2 (3 × 180/4 ) + 2cos2 (180/4) + 2sec2 (180/3) = 2sin2 (135°) + 2 cos2 (45°) + 2sec2(60°) Here, cos 45° = 1/√2 sec 60° = 1/cos⁡〖60°〗 = 1/(1/2) = 2 sin 135° = sin ( 180 – 45° ) = sin 45° = 1/√2 Putting values 2 sin2 (135°) +2 cos2 (45°) + 2sec2 (60°) = 2 × (𝟏/√𝟐)^𝟐 + 2 × (𝟏/√𝟐)^𝟐 + 2 × (2)2 = 2 [(1/√2)^2 " + " (1/√2)^2 " + 22" ] = 2 [ 1/2 + 1/2 + 4] = 2 [1 + 4] = 2 × 5 = 10 = R.H.S ∴ L.H.S. = R.H.S. Hence proved

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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