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

  1. Chapter 2 Class 12 Inverse Trigonometric Functions
  2. Serial order wise
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Misc 14 Solve tan-1 (1 โˆ’ x)/(1 + x) = 1/2 tan-1 x, (x > 0) tan-1 (1 โˆ’ x)/(1 + x) = 1/2 tan-1 x 2 tan-1 ((1 โˆ’ x)/(1 + x)) = tan-1 x We know that 2 tan-1 x = tan-1 ((๐Ÿ ๐ฑ )/(๐Ÿ โˆ’๐ฑ๐Ÿ)) 2tan-1 ((1 โˆ’ x)/(1 + x)) = tan-1 [(2 ((1 โˆ’ ๐‘ฅ)/(1 + ๐‘ฅ)))/(1 โˆ’ ((1 โˆ’ ๐‘ฅ)/(1 + ๐‘ฅ ))^2 )] = tan-1 [((2 (1 โˆ’ ๐‘ฅ))/((1 + ๐‘ฅ)))/(((1 + ๐‘ฅ)2 โˆ’ ( 1 โˆ’๐‘ฅ )2)/(1 + ๐‘ฅ )^2 )] = tan-1 [(2 (1 โˆ’ ๐‘ฅ))/((1 + ๐‘ฅ)) ร— ((1 + ๐‘ฅ))/((1 + ๐‘ฅ)2 โˆ’ (1 โˆ’ ๐‘ฅ)2)] = tan-1 [(2 (1 โˆ’ ๐‘ฅ) (1 + ๐‘ฅ))/((1 + ๐‘ฅ)2 โˆ’ (1 โˆ’ ๐‘ฅ)2)] Using (a + b) (a โ€“ b) = a2 โ€“ b2 = tan-1 [ (2 (1 โˆ’ ๐‘ฅ2) )/((1 + ๐‘ฅ + 1 โˆ’ ๐‘ฅ) (1+ ๐‘ฅ โˆ’ 1 + ๐‘ฅ) )] = tan-1 [ (2 (1 โˆ’ ๐‘ฅ2) )/((1 +1 โˆ’ ๐‘ฅ โˆ’ ๐‘ฅ) (๐‘ฅ + ๐‘ฅ โˆ’ 1 + 1) )] = tan-1 [(2 (1 โˆ’ ๐‘ฅ2))/(4 (1) (๐‘ฅ) )] = tan-1 [(1 โˆ’ ๐‘ฅ2)/2๐‘ฅ] โˆด 2tan-1 ((1 โˆ’ x)/(1 + x)) = tan-1 [(1 โˆ’ ๐‘ฅ2)/2๐‘ฅ] From (1) 2tan-1 ((1 โˆ’ x)/(1 + x) " " ) = tan-1 x Putting value tan-1 ((1 โˆ’ ๐‘ฅ2)/2๐‘ฅ) = tan-1 x (1 โˆ’ ๐‘ฅ2)/2๐‘ฅ = x 1 โ€“ x2 = x (2x) 1 โ€“ x2 = 2x2 1 โ€“ x2 โ€“ 2x2 = 0 1 โ€“ 3x2 = 0 3x2 = 1 x2 = 1/3 x = ยฑ โˆš(1/3) x = ยฑ 1/โˆš3 x = (โˆ’ 1 )/โˆš3 is not possible because Given that x > 0 ( x should be positive) Hence, x = ( 1 )/โˆš3

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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.