1. Chapter 2 Class 12 Inverse Trigonometric Functions
2. Serial order wise

Transcript

Misc 14 Solve tan-1 (1 โ x)/(1 + x) = 1/2 tan-1 x, (x > 0) tan-1 (1 โ x)/(1 + x) = 1/2 tan-1 x 2 tan-1 ((1 โ x)/(1 + x)) = tan-1 x We know that 2 tan-1 x = tan-1 ((๐ ๐ฑ )/(๐ โ๐ฑ๐)) 2tan-1 ((1 โ x)/(1 + x)) = tan-1 [(2 ((1 โ ๐ฅ)/(1 + ๐ฅ)))/(1 โ ((1 โ ๐ฅ)/(1 + ๐ฅ ))^2 )] = tan-1 [((2 (1 โ ๐ฅ))/((1 + ๐ฅ)))/(((1 + ๐ฅ)2 โ ( 1 โ๐ฅ )2)/(1 + ๐ฅ )^2 )] = tan-1 [(2 (1 โ ๐ฅ))/((1 + ๐ฅ)) ร ((1 + ๐ฅ))/((1 + ๐ฅ)2 โ (1 โ ๐ฅ)2)] = tan-1 [(2 (1 โ ๐ฅ) (1 + ๐ฅ))/((1 + ๐ฅ)2 โ (1 โ ๐ฅ)2)] Using (a + b) (a โ b) = a2 โ b2 = tan-1 [ (2 (1 โ ๐ฅ2) )/((1 + ๐ฅ + 1 โ ๐ฅ) (1+ ๐ฅ โ 1 + ๐ฅ) )] = tan-1 [ (2 (1 โ ๐ฅ2) )/((1 +1 โ ๐ฅ โ ๐ฅ) (๐ฅ + ๐ฅ โ 1 + 1) )] = tan-1 [(2 (1 โ ๐ฅ2))/(4 (1) (๐ฅ) )] = tan-1 [(1 โ ๐ฅ2)/2๐ฅ] โด 2tan-1 ((1 โ x)/(1 + x)) = tan-1 [(1 โ ๐ฅ2)/2๐ฅ] From (1) 2tan-1 ((1 โ x)/(1 + x) " " ) = tan-1 x Putting value tan-1 ((1 โ ๐ฅ2)/2๐ฅ) = tan-1 x (1 โ 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 โ(1/3) x = ยฑ 1/โ3 x = (โ 1 )/โ3 is not possible because Given that x > 0 ( x should be positive) Hence, x = ( 1 )/โ3