Q106 / 160 · TS EAMCET 2009 · Mathematics

$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \Rightarrow \frac{d y}{d x}$ is equal to

  • A
    0
  • B
    $\tan t$
  • Correct answer
    C
    1
  • D
    $\sin t \cos t$

Solution

Given, $x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ $\begin{array}{lrl} \text { and } & & y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \\ \Rightarrow & & x=\tan ^{-1} t, \\ \text { and } & y=\tan ^{-1} t \\ \therefore & y=x \Rightarrow \frac{d y}{d x}=1 \end{array}$

Hence, the correct answer is option: 'C'.

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