Question

If \(1+ (√1+x) tanx = 1+ (√1-x)\)  then \(sin4x\)  is ?

 

Answer 1

Given Equation:
We are given the equation: \[ \tan(y) = 1 + \frac{1 + x}{1 - x} \] and we substitute \( x = \cos(\theta) \) into the equation. We will simplify it step by step.

Step 1: Substitute \( x = \cos(\theta) \):
Substituting \( x = \cos(\theta) \) into the equation, we get: \[ \tan(y) = 1 + \frac{1 + \cos(\theta)}{1 - \cos(\theta)} \]

Step 2: Simplify the expression:
Now, simplifying further: \[ \tan(y) = 2 \left| \frac{1 + \cos^2(\theta)}{1 - \cos^2(\theta)} \right| \div \left| \frac{1 + \sin^2(\theta)}{1 - \sin^2(\theta)} \right| \]

Step 3: Simplify further:
We continue to simplify this expression: \[ \tan(y) = 2 \left| \frac{1 + \cos^2(\theta)}{\sin^2(\theta)} \right| \div \left| \frac{1 + \sin^2(\theta)}{\cos^2(\theta)} \right| \]

Step 4: Combine the terms:
Simplifying further, we have: \[ \tan(y) = 2 \left( \frac{\cos^4(\theta) + \cos^2(\theta)}{\sin^2(\theta)} \right) \div \left( \frac{\sin^4(\theta) + \sin^2(\theta)}{\cos^2(\theta)} \right) \]

Step 5: Simplify the final expression:
The equation simplifies to: \[ \tan(y) = \frac{2 (\cos^4(\theta) + \cos^2(\theta))}{\sin^4(\theta) + \sin^2(\theta)} \]

Step 6: Use a trigonometric identity:
This can be rewritten as: \[ \tan(y) = \frac{2 \cos(8\pi + 4\theta) \cdot \cos(8\pi - 4\theta)}{2 \sin(8\pi + 4\theta) \cdot \cos(8\pi - 4\theta)} \]

Step 7: Simplify further:
Simplifying the above equation: \[ \tan(y) = \tan(8\pi + 4\theta) \]

Step 8: Solve for \( y \):
From this equation, we can deduce that: \[ 4y = 2\pi + \theta \] Thus: \[ y = \frac{2\pi + \theta}{4} \]

Step 9: Find the relation with \( x \):
Since \( \sin(4y) = \cos(\theta) = x \), we can conclude that: \[ \sin(4y) = x \]