Answer:
To prove that \( 2 \tan^{-1} x = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \), we will use some trigonometric identities and properties of inverse trigonometric functions.
### Step-by-Step Proof
1. **Let \( \theta = \tan^{-1} x \)**
By definition, if \( \theta = \tan^{-1} x \), then:
\[ x = \tan \theta \]
2. **Double the angle**:
We need to express \( 2 \theta \) in terms of \( x \). Using the double-angle identity for tangent, we have:
\[ \tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \]
Since \( x = \tan \theta \), we substitute \( x \) for \( \tan \theta \):
\[ \tan(2\theta) = \frac{2x}{1 - x^2} \]
3. **Express \( \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \) in terms of \( \theta \)**:
We need to show that \( 2\theta = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \).
Recall the trigonometric identity:
\[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \]
Using the Pythagorean identity:
\[ \cos^2 \theta = \frac{1}{1 + \tan^2 \theta} = \frac{1}{1 + x^2} \]
\[ \sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta} = \frac{x^2}{1 + x^2} \]
Substitute these into the double-angle identity for cosine:
\[ \cos(2\theta) = \left( \frac{1}{1 + x^2} \right) - \left( \frac{x^2}{1 + x^2} \right) \]
\[ \cos(2\theta) = \frac{1 - x^2}{1 + x^2} \]
4. **Relate \( 2 \theta \) and the inverse cosine**:
Since \( \cos(2\theta) = \frac{1 - x^2}{1 + x^2} \), we can take the inverse cosine on both sides to solve for \( 2\theta \):
\[ 2\theta = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \]
5. **Substitute back \( \theta = \tan^{-1} x \)**:
Since \( \theta = \tan^{-1} x \), we substitute back:
\[ 2 \tan^{-1} x = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \]
### Conclusion
Thus, we have proved that:
\[ 2 \tan^{-1} x = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \]
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Step-by-step explanation:
To prove the given trigonometric identity, we can start by expressing the inverse trigonometric functions in terms of basic trigonometric functions using the definitions:
1. tan^(-1)(x) = y means tan(y) = x
2. cos^(-1)(x) = y means cos(y) = x
Let's consider the right side of the equation:
cos^(-1)((1-x^2)/(1+x^2))
Using the definition of cosine inverse, we have:
cos^(-1)((1-x^2)/(1+x^2)) = y
cos(y) = (1-x^2)/(1+x^2)
Now, we can rewrite the right side in terms of tan^(-1)(x):
cos(y) = (1 - x^2)/(1 + x^2)
Dividing the numerator and denominator by 1, we get:
cos(y) = (1/x^2 - 1)/(1/x^2 + 1)
cos(y) = ((1 - x^2)/(x^2))/((1 + x^2)/(x^2))
cos(y) = (1 - x^2)/(1 + x^2)
tan(y) = (1 - x^2)/(1 + x^2)
Therefore, cos^(-1)((1-x^2)/(1+x^2)) = tan^(-1)((1-x^2)/(1+x^2))
Hence, we have proved that 2tan^(-1)(x) = cos^(-1)((1-x^2)/(1+x^2)).
