solve: `2^t +2^(-t) > 2sqrt2`

Solve : (t + 2)/(3) + 1/(t + 1) = (t - 3)/(2) - (t - 1)/(6)

Solve : 2.5t + 7.3 t = 21.6 -t

int frac{ t^2 + 1}{ t^4 + 1} dt=...........a) frac{1}{sqrt2}[tan^(-1)(sqrt 2t-1)+tan^(-1)(sqrt 2t+1)]+c b) sqrt2[tan^(-1)(sqrt 2t-1)+tan^(-1)(sqrt 2t+1)]+c c) frac{1}{2sqrt2}[tan^(-1)(sqrt 2t-1)+tan^(-1)(sqrt 2t+1)]+c d) 2sqrt2[tan^(-1)(sqrt 2t-1)+tan^(-1)(sqrt 2t+1)]+c

Solve : (3t-2)/4 - (2t+3)/3 = 2/3 - t

solve the equation . (3 t - 2)/(4 ) - (2 t + 3)/(3) = 2/3 - t

Problem solving: -2(t+3)=8 , t=

Solve (dy)/(dt) = t^(2).

If x = sqrt((1 - t^2)/(1 + t^2)), y = (sqrt(1 + t^2) - sqrt(1 - t^2))/(sqrt(1 + t^2) + sqrt(1 -t^2)) then (dy)/(dx) =

The eccentricity of the hyperbola x=a/2(t+1/t), y=a/2(t-1/t) is a. sqrt(2) . b. sqrt(3) c. 2sqrt(3) d. 3sqrt(2)

x = (sin^3t)/(sqrt(cos 2t)), y = (cos^3 t)/(sqrt(cos 2t)) .