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

Transcript

Misc 14 Solve tan-1 (1 x)/(1 + x) = 1/2 tan-1 x, (x > 0) tan-1 (1 x)/(1 + x) = 1/2 tan-1 x 2 tan-1 ((1 x)/(1 + x)) = tan-1 x We know that 2 tan-1 x = tan-1 (( )/( )) 2tan-1 ((1 x)/(1 + x)) = tan-1 [(2 ((1 )/(1 + )))/(1 ((1 )/(1 + ))^2 )] = tan-1 [((2 (1 ))/((1 + )))/(((1 + )2 ( 1 )2)/(1 + )^2 )] = tan-1 [(2 (1 ))/((1 + )) ((1 + ))/((1 + )2 (1 )2)] = tan-1 [(2 (1 ) (1 + ))/((1 + )2 (1 )2)] Using (a + b) (a b) = a2 b2 = tan-1 [ (2 (1 2) )/((1 + + 1 ) (1+ 1 + ) )] = tan-1 [ (2 (1 2) )/((1 +1 ) ( + 1 + 1) )] = tan-1 [(2 (1 2))/(4 (1) ( ) )] = tan-1 [(1 2)/2 ] 2tan-1 ((1 x)/(1 + x)) = tan-1 [(1 2)/2 ] From (1) 2tan-1 ((1 x)/(1 + x) " " ) = tan-1 x Putting value tan-1 ((1 2)/2 ) = tan-1 x (1 2)/2 = x 1 x2 = x (2x) 1 x2 = 2x2 1 x2 2x2 = 0 1 3x2 = 0 3x2 = 1 x2 = 1/3 x = (1/3) x = 1/ 3 x = ( 1 )/ 3 is not possible because Given that x > 0 ( x should be positive) Hence, x = ( 1 )/ 3

Miscellaneous