2x 3x formula - Proving

Chapter 3 Class 11 Trigonometric Functions
Concept wise

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Ex 3.3, 23 Prove that tanβ‘4π₯ = (4 tanβ‘γπ₯ (1βtan2π₯)γ)/(1 β 6 tan2 π₯+tan4 π₯) Taking L.H.S. tan 4x We know that tan 2x = (2 π‘ππβ‘π₯)/(1 β π‘ππ2 π₯) Replacing x with 2x tan (2 Γ 2x) = (2 π‘ππβ‘2π₯)/(1 β π‘ππ2 2π₯) tan 4x = (2 π‘ππβ‘2π₯)/(1 β π‘ππ2 2π₯) = (2 tanβ‘2x)/(1 β tan2 2x) = 2((2 tanβ‘π₯)/(1 β π‘ππ2 π₯))/(1 β ((2 tanβ‘π₯)/(1 β π‘ππ2 π₯))^2 ) = (((4 tanβ‘π₯)/(1 β π‘ππ2 π₯)))/(1 β((2 tanβ‘π₯ )^2/(1 β π‘ππ2 π₯)^2 ) ) = (((4 tanβ‘π₯)/(1 β π‘ππ2 π₯)))/(1 β((4 γπ‘ππγ^2 π₯)/(1 β π‘ππ2 π₯)^2 ) ) Using tan 2x = (2 tanβ‘x)/(1 β tan2 x) = (((4 tanβ‘π₯)/(1 β π‘ππ2 π₯)))/((((1 β π‘ππ2 π₯)^2 β4 γπ‘ππγ^2 π₯)/(1 β π‘ππ2 π₯)^2 ) ) = (4 tanβ‘π₯)/(1 β tan2β‘π₯ ) Γ (1 β tan2β‘π₯ )^2/((1 β π‘ππ2 π₯)2 β4 tan2β‘π₯ ) = (4 tanβ‘π₯)/1 Γ ((1 β tan2β‘γπ₯)γ)/((1 β π‘ππ2 π₯)2 β 4 tan2β‘π₯ ) = (4 tanβ‘γπ₯ (1 β π‘ππ2 π₯γ))/((1 βtan2β‘π₯ )2 β 4 tanβ‘2π₯ ) Using (a β b)2 = a2 + b2 β 2ab = (4 tanβ‘γπ₯ (1 β tan2β‘π₯)γ)/(( 12+(π‘ππ2 π₯)2 β 2 Γ 1 Γ π‘ππ2 π₯) β4 π‘ππ2 π₯) = (4 tanβ‘γπ₯ (1 β tan2β‘π₯)γ)/(1 + tanβ‘γ4 π₯ β 2 tan2β‘γπ₯ β4 tan2β‘π₯ γ γ ) = (4 tanβ‘γπ₯ (1 β tan2β‘γπ₯)γ γ)/(1 + tan4β‘γπ₯ β 6 π‘ππ4 π₯γ ) = R.H.S. Hence L.H.S. = R.H.S. Hence proved