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Misc 14 - Solve tan-1 (1 - x)/(1 + x) = 1/2 tan-1 x | NCERT

Misc. 14 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Misc. 14 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3
Misc. 14 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4

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Misc 14 Solve tan-1 (1 βˆ’ x)/(1 + x) = 1/2 tan-1 x, (x > 0) tan-1 (1 βˆ’ x)/(1 + x) = 1/2 tan-1 x 2 tan-1 ((1 βˆ’ x)/(1 + x)) = tan-1 x tan-1 [(2 ((1 βˆ’ π‘₯)/(1 + π‘₯)))/(1 βˆ’ ((1 βˆ’ π‘₯)/(1 + π‘₯ ))^2 )] = tan-1 x We know that 2 tan-1 x = tan-1 ((πŸπ’™ )/(𝟏 βˆ’ 𝐱^𝟐 )) Replacing x by (1 βˆ’ π‘₯)/(1 + π‘₯) tan-1 [((2 (1 βˆ’ π‘₯))/((1 + π‘₯)))/(((1 + π‘₯)2 βˆ’ ( 1 βˆ’π‘₯ )2)/(1 + π‘₯ )^2 )] = tan-1 x tan-1 [(2 (1 βˆ’ π‘₯))/((1 + π‘₯)) Γ— ((1 + π‘₯))/((1 + π‘₯)2 βˆ’ (1 βˆ’ π‘₯)2)] = tan-1 x tan-1 [(2 (1 βˆ’ π‘₯) (1 + π‘₯))/((1 + π‘₯)2 βˆ’ (1 βˆ’ π‘₯)2)] = tan-1 x Using (a + b) (a – b) = a2 – b2 tan-1 [ (2 (1 βˆ’ π‘₯2) )/((1 + π‘₯ + 1 βˆ’ π‘₯) (1+ π‘₯ βˆ’ 1 + π‘₯) )] = tan-1 x tan-1 [ (2 (1 βˆ’ π‘₯2) )/((1 +1 βˆ’ π‘₯ βˆ’ π‘₯) (π‘₯ + π‘₯ βˆ’ 1 + 1) )] = tan-1 x tan-1 [(2 (1 βˆ’ π‘₯2))/(4 (1) (π‘₯) )] = tan-1 x tan-1 [(1 βˆ’ π‘₯2)/2π‘₯] = tan-1 x Comparing values (1 βˆ’ π‘₯2)/2π‘₯ = x 1 – x2 = 2x2 1 – x2 – 2x2 = 0 1 – 3x2 = 0 3x2 = 1 x2 = 1/3 x = Β± 1/√3 x = (βˆ’ 1 )/√3 is not possible because it is Given that x > 0 Hence, x = ( 𝟏 )/βˆšπŸ‘

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.