Ex 3.3, 7 - Prove tan (pi/4 + x) / tan (pi/4 - x) = (1 + tan x)2 - Ex 3.3

part 2 - Ex 3.3, 7 - Ex 3.3 - Serial order wise - Chapter 3 Class 11 Trigonometric Functions
part 3 - Ex 3.3, 7 - Ex 3.3 - Serial order wise - Chapter 3 Class 11 Trigonometric Functions

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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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