Ex 3.3, 23 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 16, 2024 by Teachoo
Ex 3.3
Ex 3.3, 1
Important
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Chapter 3 Class 11 Trigonometric Functions
Serial order wise
Transcript
Ex 3.3, 23 Prove that tanβ‘4π₯ = (4 tanβ‘γπ₯ (1βtan2π₯)γ)/(1 β 6 tan2 π₯+tan4 π₯) Solving L.H.S. tan 4x We know that tan 2x = (2 π‘ππβ‘π₯)/(1 β π‘ππ2 π₯) Replacing x with 2x tan (2 Γ 2x) = (2 π‘ππβ‘2π₯)/(1 β π‘ππ2 2π₯) tan 4x = (2 π‘ππβ‘2π₯)/(1 β π‘ππ2 2π₯) = (π πππ§β‘ππ±)/(π β πππ§π ππ±) = 2((π πππβ‘π)/(π β ππππ π))/(1 β ((π πππβ‘π)/(π β ππππ π))^2 ) = (((4 tanβ‘π₯)/(1 β π‘ππ2 π₯)))/(1 β((2 tanβ‘π₯ )^2/(1 β π‘ππ2 π₯)^2 ) ) = (((4 tanβ‘π₯)/(1 β π‘ππ2 π₯)))/(1 β((4 γπ‘ππγ^2 π₯)/(1 β π‘ππ2 π₯)^2 ) ) = (((4 tanβ‘π₯)/(1 β π‘ππ2 π₯)))/((((1 β π‘ππ2 π₯)^2 β4 γπ‘ππγ^2 π₯)/(1 β π‘ππ2 π₯)^2 ) ) = (π πππβ‘π)/(π β πππ§πβ‘π ) Γ (π β πππ§πβ‘π )^π/((π β ππππ π)π βπ πππ§πβ‘π ) = (4 tanβ‘π₯)/1 Γ ((1 β tan2β‘γπ₯)γ)/((1 β π‘ππ2 π₯)2 β 4 tan2β‘π₯ ) = (π πππβ‘γπ (π β ππππ πγ))/((π βπππ§πβ‘π )π β π πππβ‘ππ ) Using (a β b)2 = a2 + b2 β 2ab = (4 tanβ‘γπ₯ (1 β tan2β‘π₯)γ)/(( 12+(π‘ππ2 π₯)2 β 2 Γ 1 Γ π‘ππ2 π₯) β4 π‘ππ2 π₯) = (4 tanβ‘γπ₯ (1 β tan2β‘π₯)γ)/(1 + tanβ‘γ4 π₯ β 2 tan2β‘γπ₯ β4 tan2β‘π₯ γ γ ) = (π πππβ‘γπ (π β πππ§πβ‘γπ)γ γ)/(π + πππ§πβ‘γπ β π ππππ πγ ) = R.H.S. Hence L.H.S. = R.H.S. Hence proved
