Question

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

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Step 1. Find the value of $\frac{dy}{dx}$:

Given, $\cos x=\frac{1}{\sqrt{(1+t^2)}}$

$\Rightarrow$ $x={\cos }^{-1}\left[\frac{1}{\sqrt{(1+t^2)}}\right]$

Step 2. Put $t=\tan \theta$ in above equation,

$\begin{array}{rcl}x& =& {\cos }^{-1}\left[\frac{1}{\sqrt{(1+{\tan }^2\theta )}}\right]\\ & =& {\cos }^{-1}\left(\frac{1}{\sqrt{se{c}^2\theta }}\right)\\ & =& {\cos }^{-1}\left(\frac{1}{sec\theta }\right)\\ & =& {\cos }^{-1}\left(\cos \theta \right)\\ & =& \theta \\ & =& {\tan }^{-1}t\end{array}$

$\therefore \frac{dx}{dt}=\frac{1}{(1+t^2)}$

Also, Given $\sin y=\frac{t}{\sqrt{(1+t^2)}}$

$\Rightarrow$ $y={\sin }^{-1}\left[\frac{t}{\sqrt{(1+t^2)}}\right]$

Step 3. Put $\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\theta }$ in above equation,

$\begin{array}{rcl}y& =& {\sin }^{-1}\left[\frac{\tan \theta }{\sqrt{(1+{\tan }^2\theta )}}\right]\\ & =& {\sin }^{-1}\left(\frac{\tan \theta }{\sqrt{se{c}^2\theta }}\right)\\ & =& {\sin }^{-1}\left(\frac{\tan \theta }{sec\theta }\right)\\ & =& {\sin }^{-1}\left(\sin \theta \right)\\ & =& \theta \\ & =& {\tan }^{-1}t\end{array}$

$\therefore \frac{dy}{dt}=\frac{1}{(1+t^2)}$

Step 4. Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$, we get

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{(dy/dt)}{(dx/dt)}\\ & =& \frac{dy}{dt}\times \frac{dt}{dx}\\ & =& \frac{1}{\left(1+t^2\right)}\times \frac{\left(1+t^2\right)}{1}\end{array}$

$\mathbf{\therefore }\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Hence, Option ‘A’ is Correct.