Question

If$\tan y=\frac{2t}{(1-t^2)}$ and $\sin x=\frac{2t}{(1+t^2)}$, then $\frac{dy}{dx}$ =

• A. $\frac{2}{\left(1-t^2\right)}$
• B. $\frac{1}{\left(1+t^2\right)}$
• C. $1$
• D. $2$

Answer 1

The correct option is C

$1$

Explanation for Correct Option:

Given,

$\sin \mathrm{x}=\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\mathrm{and}\tan \mathrm{y}=\frac{2\mathrm{t}}{1-{\mathrm{t}}^2}$

Step 1: Differentiate $\sin \mathrm{x}$ with respect to t

$$\begin{gathered} \cos x=\frac{dx}{dt}=\frac{(1+t^2)\left(2\right)-2t(0+2t)}{{\left(1+t^2\right)}^2} \\ \cos x=\frac{dx}{dt}=\frac{2+2t^2-4t^2}{{\left(1+t^2\right)}^2} \\ \cos x=\frac{dx}{dt}=\frac{2-2t^2}{{\left(1+t^2\right)}^2} \end{gathered}$$

$$\begin{gathered} \frac{dx}{dt}=\frac{2\left(1-t^2\right)}{{\left(1+t^2\right)}^2}\times \frac{1+t^2}{\sqrt{{\left(1+t^2\right)}^2-4t^2}}\left[\because \cos x=\sqrt{1-{\sin }^{2}x}\right] \\ \frac{dx}{dt}=\frac{2\left(1-t^2\right)}{{\left(1+t^2\right)}^2}\times \frac{1+t^2}{\sqrt{1+t^4+2t^2-4t^2}} \\ \frac{dx}{dt}=\frac{2\left(1-t^2\right)}{{\left(1+t^2\right)}^2}\times \frac{1+t^2}{\sqrt{{\left(1+t^2\right)}^2}}=\frac{2}{1+t^2}....\left(\mathrm{i}\right) \end{gathered}$$

Step 2: Differentiate $\tan \mathrm{x}$ with respect to t

$$\begin{gathered} \tan y=\frac{2t}{1-t^2} \\ se{c}^{2}y\frac{dy}{dt}=\frac{(1-t^2)\left(2\right)-2t(0-2t)}{{\left(1-t^2\right)}^2}=\frac{2+2t^2}{{\left(1-t^2\right)}^2} \\ \frac{dy}{dt}=\frac{2(1+t^{2)}}{{(1-t^2)}^2}\times \frac{1}{\left[1+{\left[{\displaystyle \frac{2t}{1-t^2}}\right]}^2\right]}\left[\because se{c}^{2}x={\tan }^{2}x+1\right] \\ \frac{dy}{dt}=\frac{2(1+t^{2)}}{{(1-t^2)}^2}\times \frac{{(1-t^2)}^2}{{(1-t^2)}^2+4t^2}=\frac{2(1-t^{2)}}{{(1-t^2)}^2} \\ \frac{dy}{dt}=\frac{2}{1+t^2}....\left(\mathrm{ii}\right) \end{gathered}$$

Step 3: Differentiate equations (i) and (ii) with respect to x

$\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{{\displaystyle \frac{2}{1-t^2}}}{{\displaystyle \frac{2}{1-t^2}}}=1$

Hence Option (3) is correct.