Question

Consider the following statements
Assertion (A): For \(x \in \mathbb{R} - \{1\}\), \(\frac{d}{dx}\left(\tan^{-1}\left(\frac{1+x}{1-x}\right)\right) = \frac{d}{dx}(\tan^{-1}x)\)
Reason (R): For \(x<1\), \(\tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1}x\),
for \(x>1\), \(\tan^{-1}\left(\frac{1+x}{1-x}\right) = -\frac{3\pi}{4} + \tan^{-1}x\)
The correct answer is

• A. Both (A) and (R) are true, (R) is the correct explanation of (A)
• B. Both (A) and (R) are true, (R) is not the correct explanation of (A)
• C. (A) is true, but (R) is false
• D. (A) is false, but (R) is true

Hint:

Inverse trigonometric identities often have conditional branches (e.g., \(xy<1\) vs \(xy>1\)). The derivative usually remains the same functional form because the branches differ only by a constant.

Answer 1

The Correct Option is A

Step 1: Analyze Reason (R):
The identity for \(\tan^{-1} x + \tan^{-1} y\) depends on the product \(xy\). Let \(y=1\). Then \(\tan^{-1}(1) + \tan^{-1}(x) = \tan^{-1}\left(\frac{1+x}{1-x}\right)\) provided \(1 \cdot x<1\) (i.e., \(x<1\)). So for \(x<1\), \(\tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1}x\). If \(x>1\), then \(1 \cdot x>1\). The formula becomes \(\tan^{-1} x + \tan^{-1} 1 = \pi + \tan^{-1}\left(\frac{x+1}{1-x}\right)\). Rearranging: \(\tan^{-1}\left(\frac{1+x}{1-x}\right) = \tan^{-1}x + \frac{\pi}{4} - \pi = \tan^{-1}x - \frac{3\pi}{4}\). Thus, Reason (R) is correct.
Step 2: Analyze Assertion (A):
Differentiating the expressions given in R: For \(x<1\): \(\frac{d}{dx}\left(\frac{\pi}{4} + \tan^{-1}x\right) = 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}\). For \(x>1\): \(\frac{d}{dx}\left(-\frac{3\pi}{4} + \tan^{-1}x\right) = 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}\). In both cases, the derivative is equal to \(\frac{d}{dx}(\tan^{-1}x)\). Assertion (A) is true.
Step 3: Link A and R:
The Assertion follows directly from the piecewise definitions provided in the Reason, as the derivative of the constant terms (\(\pi/4\) and \(-3\pi/4\)) is zero. Hence, R explains A.