Let t (0, 1) and (0, 4 ) . If x = cosec ⁻¹ ( 1+t^2 2t ) , y = ⁻¹ ( 1-t^2 t ) and dy dx = f(t) , then the value of f( ) is

Step-by-step solution

Given x = cosec ⁻¹ ( 1+t^2 2t ) = ⁻¹ ( 2t 1+t^2 ) Since t (0, 1) , we can write x = 2 ⁻¹t Differentiating with respect to t , we get: dx dt = 2 1+t^2 Also given y = ⁻¹ ( 1-t^2 t ) Let t = , then 1-t^2 t = = So, y = ⁻¹( ) = = ⁻¹t Differentiating with respect to t , we get: dy dt = 1 1-t^2 Now, f(t) = dy dx = dy dt dx dt = 1 1-t^2 2 1+t^2 = 1+t^2 2 1-t^2 Substituting t = : f( ) = 1+ ^2 2 1- ^2 f( ) = ^2 2 ^2 - ^2 ^2 f( ) = ^2 2 2 f( ) = ^2 2 2 = 2 2 Answer: 2 2