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Ex 3.3, 4 - Chapter 3 Class 11 Trigonometric Functions
Last updated at June 22, 2023 by Teachoo
Ex 3.3
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Ex 3.3, 4 You are here
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Chapter 3 Class 11 Trigonometric Functions
Serial order wise
Transcript
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
