If `cos x =1/sqrt(1+t^(2))`, and `sin y = t/sqrt(1+t^(2))`, then `(dy)/(dx)` =

Options

A. `-1`
B. `1/(1+t^(2))`
C. `(1-t)/(1+t^(2))`
D. 1

Answer

To find \(\frac{dy}{dx}\) given the equations \( \cos x = \frac{1}{\sqrt{1+t^2}} \) and \( \sin y = \frac{t}{\sqrt{1+t^2}} \), we can use implicit differentiation and the chain rule.

### Step-by-Step Solution:

1. **Differentiate \( \cos x \) with respect to \( t \)**:
\[ \frac{d}{dt}(\cos x) = -\sin x \cdot \frac{dx}{dt} \]
From the equation \( \cos x = \frac{1}{\sqrt{1+t^2}} \), we differentiate the right-hand side:
\[ \frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right) = -\frac{1}{2}(1+t^2)^{-3/2} \cdot 2t = -\frac{t}{(1+t^2)^{3/2}} \]
Thus, we have:
\[ -\sin x \cdot \frac{dx}{dt} = -\frac{t}{(1+t^2)^{3/2}} \]

2. **Differentiate \( \sin y \) with respect to \( t \)**:
\[ \frac{d}{dt}(\sin y) = \cos y \cdot \frac{dy}{dt} \]
From the equation \( \sin y = \frac{t}{\sqrt{1+t^2}} \), we differentiate the right-hand side:
\[ \frac{d}{dt}\left(\frac{t}{\sqrt{1+t^2}}\right) = \frac{\sqrt{1+t^2} \cdot 1 - t \cdot \frac{1}{2}(1+t^2)^{-1/2} \cdot 2t}{1+t^2} = \frac{(1+t^2) - t^2}{(1+t^2)^{3/2}} = \frac{1}{(1+t^2)^{3/2}} \]
Thus, we have:
\[ \cos y \cdot \frac{dy}{dt} = \frac{1}{(1+t^2)^{3/2}} \]

3. **Express \(\frac{dy}{dt}\) in terms of \(\frac{dx}{dt}\)**:
From the first equation:
\[ \frac{dx}{dt} = \frac{t}{\sin x (1+t^2)^{3/2}} \]
From the second equation:
\[ \frac{dy}{dt} = \frac{1}{\cos y (1+t^2)^{3/2}} \]

4. **Use the chain rule to find \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1}{\cos y (1+t^2)^{3/2}}}{\frac{t}{\sin x (1+t^2)^{3/2}}} = \frac{\sin x}{t \cos y} \]

5. **Substitute \(\sin x\) and \(\cos y\)**:
From the definitions:
\[ \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \left(\frac{1}{\sqrt{1+t^2}}\right)^2} = \sqrt{\frac{t^2}{1+t^2}} = \frac{t}{\sqrt{1+t^2}} \]
And since \( \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - \left(\frac{t}{\sqrt{1+t^2}}\right)^2} = \frac{1}{\sqrt{1+t^2}} \).

6. **Final expression for \(\frac{dy}{dx}\)**:
Substitute these into the expression:
\[ \frac{dy}{dx} = \frac{\frac{t}{\sqrt{1+t^2}}}{t \cdot \frac{1}{\sqrt{1+t^2}}} = 1 \]

### Final Answer:
\[ \frac{dy}{dx} = 1 \]