$\lim_{x \to 1/\sqrt{2}} \frac{x - \cos(\sin^{-1} x)}{1 - \tan(\sin^{-1} x)}$ ની કિંમત શોધો.

  • A
    $\frac{1}{\sqrt{2}}$
  • B
    $-\frac{1}{\sqrt{2}}$
  • C
    $\sqrt{2}$
  • D
    $-\sqrt{2}$

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જો $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ હોય, તો $\tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)=$

વિધેય $f(x) = \cos^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x - 3\sin x) \right\} + \sin^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x + 3\sin x) \right\}$ નું $x = \frac{3}{4}$ આગળ $x$ ની સાપેક્ષે વિકલન શોધો.

સરવાળો $\sum\limits_{n = 1}^\infty {{\cot }^{ - 1}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ ની કિંમત શોધો.

ધારો કે $Z$ એ પૂર્ણાંક સંખ્યાઓનો ગણ છે. તો યાદી-$I$ માંની વસ્તુઓને યાદી-$II$ માંની વસ્તુઓ સાથે જોડો.
યાદી-$I$ યાદી-$II$
$A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$ $I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$
$B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$ $II$. $k \pi \pm 1, k \in Z$
$C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$ $III$. $\frac{3}{2}$
$D$. $\sin ^{-1}|\sin x|=\sqrt{\sin ^{-1}|\sin x|} \Rightarrow x \in$ $IV$. $\frac{3 \pi}{8}$
$V$. $\frac{\pi}{2}$

સાચી જોડ પસંદ કરો:

સમીકરણ $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$ ના

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