જો $\frac{1}{2} \leq x \leq 1$ હોય, તો $\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)$ ની કિંમત શોધો.

  • A
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{3}$
  • C
    $\pi$
  • D
    $0$

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કિંમત શોધો: $\operatorname{cosec}^{-1}\left[\left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}+\cos \frac{\alpha}{2} \cdot \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right]$

$\frac{d}{dx}(\sin^{-1}(3x - 4x^3)) = $

જો $\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ જ્યાં $a, x \in(0,1)$,તો $x$ ની કિંમત શોધો.

જો $\sin^{-1}(x - 2) + \cos^{-1}(x) + \tan^{-1}(x + 2) + \cot^{-1}(x + 4) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$ હોય, તો $\cos(2 \csc^{-1}\sqrt{k - 1}) = \dots$

જો $\theta = \sec^{-1}(\cosh u)$ હોય,તો $u =$

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