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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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‘₯) Solving 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⁑𝒙)/(𝟏 βˆ’ π’•π’‚π’πŸ 𝒙))/(1 βˆ’ ((𝟐 𝒕𝒂𝒏⁑𝒙)/(𝟏 βˆ’ π’•π’‚π’πŸ 𝒙))^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 ) ) = (((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 ) ) = (πŸ’ 𝒕𝒂𝒏⁑𝒙)/(𝟏 βˆ’ π­πšπ§πŸβ‘π’™ ) Γ— (𝟏 βˆ’ π­πšπ§πŸβ‘π’™ )^𝟐/((𝟏 βˆ’ π’•π’‚π’πŸ 𝒙)𝟐 βˆ’πŸ’ π­πšπ§πŸβ‘π’™ ) = (4 tan⁑π‘₯)/1 Γ— ((1 βˆ’ tan2⁑〖π‘₯)γ€—)/((1 βˆ’ π‘‘π‘Žπ‘›2 π‘₯)2 βˆ’ 4 tan2⁑π‘₯ ) = (πŸ’ 𝒕𝒂𝒏⁑〖𝒙 (𝟏 βˆ’ π’•π’‚π’πŸ 𝒙〗))/((𝟏 βˆ’π­πšπ§πŸβ‘π’™ )𝟐 βˆ’ πŸ’ π’•π’‚π’β‘πŸπ’™ ) 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€— ) = (πŸ’ 𝒕𝒂𝒏⁑〖𝒙 (𝟏 βˆ’ π­πšπ§πŸβ‘γ€–π’™)γ€— γ€—)/(𝟏 + π­πšπ§πŸ’β‘γ€–π’™ βˆ’ πŸ” π’•π’‚π’πŸ’ 𝒙〗 ) = R.H.S. Hence L.H.S. = R.H.S. Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.