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

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

NTA AbhyasNTA Abhyas 2020

Solution:

$\because xy=1\Rightarrow y=\frac{1}{x}$
$\frac{d y}{d x}=\frac{- 1}{x^{2}}$
at $x=\frac{1}{\sqrt{3}},\frac{d y}{d x}=-3$