Question:
The equation \(\tan^{-1}(1+x) + \tan^{-1}(1-x) = \frac{\pi}{2}\) is satisfied by:
समीकरण \(\tan^{-1}(1+x) + \tan^{-1}(1-x) = \frac{\pi}{2}\) किससे संतुष्ट होता है:
Options:
- \(x = 1\)
- \(x = -1\)
- \(x = 0\)
- \(x = 1/2\)
- \(x = 1\)
- \(x = -1\)
- \(x = 0\)
- \(x = 1/2\)
Correct Answer: c)
Explanation:
We know that \(\tan^{-1} A + \tan^{-1} B = \frac{\pi}{2}\) if and only if \(AB = 1\) (where \(A, B > 0\)):
\[(1+x)(1-x) = 1 \implies 1 – x^2 = 1 \implies x^2 = 0 \implies x = 0\]
Checking \(x = 0\):
\(\tan^{-1}(1) + \tan^{-1}(1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}\), which satisfies the equation.
Hence, the correct option is (c).
हम जानते हैं कि \(\tan^{-1} A + \tan^{-1} B = \frac{\pi}{2}\) तब होता है जब \(A \cdot B = 1\) हो:
\[(1+x)(1-x) = 1 \implies 1 – x^2 = 1 \implies x^2 = 0 \implies x = 0\]
\(x = 0\) रखने पर:
\(\tan^{-1}(1) + \tan^{-1}(1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}\), जो समीकरण को संतुष्ट करता है।
अतः सही विकल्प (c) है।