$\cot ^{ - 1}\left[ \frac{\sqrt {1 - \sin x} + \sqrt {1 + \sin x}}{\sqrt {1 - \sin x} - \sqrt {1 + \sin x}} \right] = $

  • A
    $\pi - x$
  • B
    $2\pi - x$
  • C
    $\frac{x}{2}$
  • D
    $\pi - \frac{x}{2}$

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$\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=0$ હોય,તો $x=$ . . . . . . .

$\cot ^{-1}\left(2 \cdot 1^2\right)+\cot ^{-1}\left(2 \cdot 2^2\right)+\cot ^{-1}\left(2 \cdot 3^2\right)+\ldots \ldots \ldots \infty =$

$S = \tan^{-1}\left( \frac{1}{n^2 + n + 1} \right) + \tan^{-1}\left( \frac{1}{n^2 + 3n + 3} \right) + \dots + \tan^{-1}\left( \frac{1}{1 + (n + 19)(n + 20)} \right)$ હોય,તો $\tan S$ ની કિંમત શોધો.

ધારો કે $S$ એ સમીકરણ $\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^2}\right)=\pi$ ના તમામ ઉકેલોનો ગણ છે,જ્યાં $x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$. તો $\sum_{x \in S} 2 \sin ^{-1}\left(x^2-1\right)$ ની કિંમત શોધો.

જો $\sum_{n=1}^{2026} \tan^{-1}(\frac{1}{n^2+n+1}) = \tan^{-1}(1 - \frac{1}{x})$, જ્યાં $x \neq 0$, તો $x = $

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