1. Class 11
2. Important Question for exams Class 11

Transcript

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 = (2 tanβ‘2x)/(1 β tan2 2x) tan 4x = 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 ) ) = (((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‘γπ₯)2γ)/((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)/(( (1)2+(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)2+(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

Class 11
Important Question for exams Class 11

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.