यदि $\sin ^{-1} x < \cos ^{-1} x$ है, तो

  • A
    $-1 \leq x < \frac{1}{\sqrt{2}}$
  • B
    $-\sqrt{3} \leq x < -1$
  • C
    $\frac{1}{\sqrt{2}} < x \leq 1$
  • D
    $1 < x < \sqrt{3}$

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निम्नलिखित कथनों पर विचार करें:
अभिकथन $(A)$: जब $x, y, z$ धनात्मक संख्याएँ हैं, तब $\operatorname{Tan}^{-1}\left(\sqrt{\frac{x(x+y+z)}{y z}}\right)+\operatorname{Tan}^{-1}\left(\sqrt{\frac{y(x+y+z)}{x z}}\right)+\operatorname{Tan}^{-1}\left(\sqrt{\frac{z(x+y+z)}{x y}}\right) = \pi$
कारण $(R)$: $\operatorname{Tan}^{-1} a + \operatorname{Tan}^{-1} b = \operatorname{Tan}^{-1}\left(\frac{a+b}{1-ab}\right)$ यदि $a > 0$ और $b > 0$ और $ab < 1$ है।

$\cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right) = $

यदि $0 < x < 1$ है,तो $\sqrt{1 + x^2} [\{x \cos (\cot^{-1} x) + \sin (\cot^{-1} x)\} ^2 - 1]^{\frac{1}{2}} =$ क्या होगा?

$\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)$ का मान है

यदि $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$ है,तो $x$ के मान ज्ञात कीजिए।

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