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

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

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

  1. Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)
  2. Concept wise

Transcript

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 tan-1 [(2 ((1 โˆ’ ๐‘ฅ)/(1 + ๐‘ฅ)))/(1 โˆ’ ((1 โˆ’ ๐‘ฅ)/(1 + ๐‘ฅ ))^2 )] = tan-1 x We know that 2 tan-1 x = tan-1 ((๐Ÿ๐’™ )/(๐Ÿ โˆ’ ๐ฑ^๐Ÿ )) Replacing x by (1 โˆ’ ๐‘ฅ)/(1 + ๐‘ฅ) 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 x 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 x tan-1 [(2 (1 โˆ’ ๐‘ฅ) (1 + ๐‘ฅ))/((1 + ๐‘ฅ)2 โˆ’ (1 โˆ’ ๐‘ฅ)2)] = tan-1 x 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 x tan-1 [ (2 (1 โˆ’ ๐‘ฅ2) )/((1 +1 โˆ’ ๐‘ฅ โˆ’ ๐‘ฅ) (๐‘ฅ + ๐‘ฅ โˆ’ 1 + 1) )] = tan-1 x tan-1 [(2 (1 โˆ’ ๐‘ฅ2))/(4 (1) (๐‘ฅ) )] = tan-1 x tan-1 [(1 โˆ’ ๐‘ฅ2)/2๐‘ฅ] = tan-1 x Comparing values (1 โˆ’ ๐‘ฅ2)/2๐‘ฅ = x 1 โ€“ x2 = 2x2 1 โ€“ x2 โ€“ 2x2 = 0 1 โ€“ 3x2 = 0 3x2 = 1 x2 = 1/3 x = ยฑ 1/โˆš3 x = (โˆ’ 1 )/โˆš3 is not possible because it is Given that x > 0 Hence, x = ( ๐Ÿ )/โˆš๐Ÿ‘

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.