If ` e^(x) = (sqrt(1+t) - sqrt(1-t))/( sqrt(1+t) + sqrt(1-t)) and tan ""(y )/(2) = sqrt((1-t)/(1+t)) " then " (dy)/(dx) "at" t= 1/2 ` is

Options

A. `1//2`
B. `-1//2`
C. `0`
D. None

Answer

To solve the problem step by step, we need to find \(\frac{dy}{dx}\) at \(t = \frac{1}{2}\) given the equations:

\[ e^x = \frac{\sqrt{1+t} - \sqrt{1-t}}{\sqrt{1+t} + \sqrt{1-t}} \]

and

\[ \tan\left(\frac{y}{2}\right) = \sqrt{\frac{1-t}{1+t}}. \]

### Step 1: Substitute \(t = \cos(2\theta)\)

We substitute \(t\) with \(\cos(2\theta)\). Then we have:

\[ 1 + t = 1 + \cos(2\theta) = 2\cos^2(\theta) \]
\[ 1 - t = 1 - \cos(2\theta) = 2\sin^2(\theta) \]

### Step 2: Rewrite \(e^x\)

Substituting these into the equation for \(e^x\):

\[ e^x = \frac{\sqrt{2\cos^2(\theta)} - \sqrt{2\sin^2(\theta)}}{\sqrt{2\cos^2(\theta)} + \sqrt{2\sin^2(\theta)}} \]

This simplifies to:

\[ e^x = \frac{\sqrt{2}(\cos(\theta) - \sin(\theta))}{\sqrt{2}(\cos(\theta) + \sin(\theta))} = \frac{\cos(\theta) - \sin(\theta)}{\cos(\theta) + \sin(\theta)}. \]

### Step 3: Rewrite \(\tan\left(\frac{y}{2}\right)\)

Now substituting into the equation for \(\tan\left(\frac{y}{2}\right)\):

\[ \tan\left(\frac{y}{2}\right) = \sqrt{\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}} = \sqrt{\frac{2\sin^2(\theta)}{2\cos^2(\theta)}} = \tan(\theta). \]

Thus, we have:

\[ \frac{y}{2} = \theta \implies y = 2\theta. \]

### Step 4: Differentiate \(y\)

Now we differentiate \(y\) with respect to \(t\):

\[ \frac{dy}{dt} = 2 \frac{d\theta}{dt}. \]

### Step 5: Find \(\frac{dx}{dt}\)

To find \(\frac{dx}{dt}\), we need to differentiate \(e^x\):

Using the chain rule:

\[ e^x \frac{dx}{dt} = \frac{d}{dt}\left(\frac{\cos(\theta) - \sin(\theta)}{\cos(\theta) + \sin(\theta)}\right). \]

### Step 6: Find \(\frac{d\theta}{dt}\)

Using the relation \(t = \cos(2\theta)\):

\[ \frac{dt}{d\theta} = -2\sin(2\theta) = -4\sin(\theta)\cos(\theta). \]

Thus,

\[ \frac{d\theta}{dt} = -\frac{1}{4\sin(\theta)\cos(\theta)}. \]

### Step 7: Substitute \(t = \frac{1}{2}\)

At \(t = \frac{1}{2}\):

\[ \cos(2\theta) = \frac{1}{2} \implies 2\theta = \frac{\pi}{3} \implies \theta = \frac{\pi}{6}. \]

### Step 8: Calculate \(e^x\) and \(\frac{dy}{dx}\)

Substituting \(\theta = \frac{\pi}{6}\):

\[ e^x = \frac{\cos(\frac{\pi}{6}) - \sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6})} = \frac{\frac{\sqrt{3}}{2} - \frac{1}{2}}{\frac{\sqrt{3}}{2} + \frac{1}{2}} = \frac{\frac{\sqrt{3}-1}{2}}{\frac{\sqrt{3}+1}{2}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}. \]

### Step 9: Final Calculation of \(\frac{dy}{dx}\)

Using the chain rule:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \cdot \left(-\frac{1}{4\sin(\frac{\pi}{6})\cos(\frac{\pi}{6})}\right)}{e^x \cdot \frac{d}{dt}\left(\frac{\cos(\theta) - \sin(\theta)}{\cos(\theta) + \sin(\theta)}\right)}. \]

### Step 10: Evaluate at \(t = \frac{1}{2}\)

Now, substitute the values and simplify to find \(\frac{dy}{dx}\) at \(t = \frac{1}{2}\).