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Chapter 3: Problem 75

Prove the following identities and give the values of \(x\) for which they are true. $$\tan \left(2 \tan ^{-1} x\right)=\frac{2 x}{1-x^{2}}$$

Short Answer

Expert verified
Question: Prove the identity \(\tan\left(2\tan^{-1}x\right) = \frac{2x}{1-x^2}\) and state the values of \(x\) for which the identity is true. Answer: The identity \(\tan\left(2\tan^{-1}x\right) = \frac{2x}{1-x^2}\) is proven using the double angle formula for tangent. The identity is true for all values of \(x\) except \(x \neq \pm 1\).

Step by step solution

01

Recall the double-angle formula for tangent

The double-angle formula for tangent is given by: $$\tan(2A) = \frac{2\tan A}{1 - (\tan A)^2}$$
02

Set up the given expression with the double-angle formula

The given expression is: $$\tan\left(2\tan^{-1}x\right) = \frac{2x}{1-x^2}$$ Let \(A = \tan^{-1}x\). Then, the left side of the expression becomes: $$\tan(2A)$$ We will apply the double-angle formula for tangent from Step 1.
03

Application of the double-angle formula for tangent

Substitute \(A = \tan^{-1}x\) in the double-angle formula for tangent: $$\tan(2\tan^{-1}x) = \frac{2\tan(\tan^{-1}x)}{1 - (\tan(\tan^{-1}x))^2}$$
04

Simplify with the inverse trigonometric function property

Since \(A = \tan^{-1}x\), we know that \(\tan A = x\). Now, substitute this into the previous expression: $$\tan(2\tan^{-1}x) = \frac{2x}{1 - x^2}$$
05

Determine the values of x for which the identity is true

The only restrictions on the domain of this identity come from the denominator of the expression. The denominator should not be equal to zero, as division by zero is undefined. Hence, $$1 - x^2 \neq 0$$ Solving for \(x\): $$ x^2 \neq 1$$ $$ x \neq \pm 1$$ Therefore, the identity holds true for all \(x\) such that \(x \neq \pm 1\).

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