MCQ
સમીકરણ ${\tan ^{ - 1}}(1 + x)$ $ + {\tan ^{ - 1}}(1 - x)$ $ = \frac{\pi }{2}$ નો ઉકેલ મેળવો.
- A$x = 1$
- B$x = - 1$
- ✓$x = 0$
- D$x = \pi $
==> ${\tan ^{ - 1}}(1 + x) = \frac{\pi }{2} - {\tan ^{ - 1}}(1 - x)$
==> ${\tan ^{ - 1}}(1 + x) = {\cot ^{ - 1}}(1 - x)$
==> ${\tan ^{ - 1}}(1 + x) = {\tan ^{ - 1}}\left( {\frac{1}{{1 - x}}} \right)$
==> $1 + x = \frac{1}{{1 - x}} \Rightarrow 1 - {x^2} = 1 \Rightarrow x = 0$.
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