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Ex 3.3, 7 Prove that: (tan"(" πœ‹/4 " + " π‘₯")" )/(tan"(" Ο€/4 " βˆ’ " π‘₯")" ) = ((1+ tan" " π‘₯)/(1βˆ’ tan" " π‘₯))^2 Solving L.H.S. (tan⁑ (πœ‹/4 + π‘₯) )/tan⁑(πœ‹/4 βˆ’ π‘₯) Numerator Numerator is of form tan (x + y) tan (x + y) = (π‘‘π‘Žπ‘›" " π‘₯ + π‘‘π‘Žπ‘›β‘π‘¦)/(1 βˆ’ π‘‘π‘Žπ‘› π‘₯ π‘‘π‘Žπ‘›β‘π‘¦ ) Putting x = 𝝅/πŸ’ , y = x tan (Ο€/4 + x) = (tan Ο€/4 + tan⁑x)/(1βˆ’ tan Ο€/4 tan⁑π‘₯ ) Now, tan πœ‹/4 = tan 45Β° = 1 tan (𝝅/πŸ’ + x) = (𝟏 + 𝒕𝒂𝒏⁑𝒙)/(πŸβˆ’ 𝒕𝒂𝒏⁑𝒙 ) Denominator Denominator is of form tan (x – y) tan (x – y) = (π‘‘π‘Žπ‘›" " π‘₯ βˆ’ π‘‘π‘Žπ‘›β‘π‘¦)/(1 + π‘‘π‘Žπ‘› π‘₯ π‘‘π‘Žπ‘›β‘π‘¦ ) Putting x = πœ‹/4 , y = x tan (Ο€/4 – x) = (tan Ο€/4 βˆ’ tan⁑x)/(1 + tan Ο€/4 tan⁑π‘₯ ) Now, tan πœ‹/4 = tan 45Β° = 1 tan (𝝅/πŸ’ – x) = (𝟏 βˆ’ 𝒕𝒂𝒏⁑𝒙)/(𝟏 + 𝒕𝒂𝒏⁑𝒙 ) Solving L.H.S tan⁑(πœ‹/4 + π‘₯)/tan⁑( πœ‹/4 βˆ’π‘₯) = ((𝟏 + 𝒕𝒂𝒏⁑𝒙)/(πŸβˆ’ 𝒕𝒂𝒏⁑𝒙 ))/((𝟏 βˆ’ 𝒕𝒂𝒏⁑𝒙)/(𝟏 + 𝒕𝒂𝒏⁑𝒙 )) = (1 + π‘‘π‘Žπ‘›β‘π‘₯)/(1βˆ’ π‘‘π‘Žπ‘›β‘π‘₯ ) Γ— (1 + π‘‘π‘Žπ‘›β‘π‘₯)/(1βˆ’ π‘‘π‘Žπ‘›β‘π‘₯ ) = (𝟏 + 𝒕𝒂𝒏⁑𝒙 )𝟐/(πŸβˆ’ 𝒕𝒂𝒏⁑𝒙 )^𝟐 = 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo