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Ex 3.3, 23 - Prove tan 4x = 4 tan x (1 - tan2 x) / 1 - 6tan2x

Ex 3.3, 23 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Ex 3.3, 23 - Chapter 3 Class 11 Trigonometric Functions - Part 3
Ex 3.3, 23 - Chapter 3 Class 11 Trigonometric Functions - Part 4

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Ex 3.3, 23 Prove that tan⁑4π‘₯ = (4 tan⁑〖π‘₯ (1βˆ’tan2π‘₯)γ€—)/(1 βˆ’ 6 tan2 π‘₯+tan4 π‘₯) Taking L.H.S. tan 4x We know that tan 2x = (2 π‘‘π‘Žπ‘›β‘π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯) Replacing x with 2x tan (2 Γ— 2x) = (2 π‘‘π‘Žπ‘›β‘2π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 2π‘₯) tan 4x = (2 π‘‘π‘Žπ‘›β‘2π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 2π‘₯) = (2 tan⁑2x)/(1 βˆ’ tan2 2x) = 2((2 tan⁑π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯))/(1 βˆ’ ((2 tan⁑π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯))^2 ) = (((4 tan⁑π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)))/(1 βˆ’((2 tan⁑π‘₯ )^2/(1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)^2 ) ) = (((4 tan⁑π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)))/(1 βˆ’((4 γ€–π‘‘π‘Žπ‘›γ€—^2 π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)^2 ) ) Using tan 2x = (2 tan⁑x)/(1 βˆ’ tan2 x) = (((4 tan⁑π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)))/((((1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)^2 βˆ’4 γ€–π‘‘π‘Žπ‘›γ€—^2 π‘₯)/(1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)^2 ) ) = (4 tan⁑π‘₯)/(1 βˆ’ tan2⁑π‘₯ ) Γ— (1 βˆ’ tan2⁑π‘₯ )^2/((1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)2 βˆ’4 tan2⁑π‘₯ ) = (4 tan⁑π‘₯)/1 Γ— ((1 βˆ’ tan2⁑〖π‘₯)γ€—)/((1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)2 βˆ’ 4 tan2⁑π‘₯ ) = (4 tan⁑〖π‘₯ (1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯γ€—))/((1 βˆ’tan2⁑π‘₯ )2 βˆ’ 4 tan⁑2π‘₯ ) Using (a – b)2 = a2 + b2 – 2ab = (4 tan⁑〖π‘₯ (1 βˆ’ tan2⁑π‘₯)γ€—)/(( 12+(π‘‘π‘Žπ‘›2 π‘₯)2 βˆ’ 2 Γ— 1 Γ— π‘‘π‘Žπ‘›2 π‘₯) βˆ’4 π‘‘π‘Žπ‘›2 π‘₯) = (4 tan⁑〖π‘₯ (1 βˆ’ tan2⁑π‘₯)γ€—)/(1 + tan⁑〖4 π‘₯ βˆ’ 2 tan2⁑〖π‘₯ βˆ’4 tan2⁑π‘₯ γ€— γ€— ) = (4 tan⁑〖π‘₯ (1 βˆ’ tan2⁑〖π‘₯)γ€— γ€—)/(1 + tan4⁑〖π‘₯ βˆ’ 6 π‘‘π‘Žπ‘›4 π‘₯γ€— ) = R.H.S. Hence L.H.S. = R.H.S. Hence proved

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