If x=sec t+tan â¡ t and y=sec t-tan â¡ t , where t is a parameter, then the value of (d y/d x) when x=(1/√3) is

Question

If x=sec t+tan â¡ t and y=sec t-tan â¡ t , where t is a parameter, then the value of (d y/d x) when x=(1/√3) is (A) 0 (B) -3 (C) √3 (D) (1/√3)

Tardigrade Admin · Accepted Answer

∵ xy=1⇒ y=(1/x) (d y/d x)=(- 1/x2) at x=(1/√3),(d y/d x)=-3

Q. If $x = sec t + t an t$ and $y = sec t - t an t$ , where $t$ is a parameter, then the value of $\frac{d y}{d x}$ when $x = \frac{1}{3}$ is