$\sec \left(\tan ^{-1} \frac{y}{2}\right) = $

  • A
    $\sqrt{\frac{4+y^2}{2}}$
  • B
    $\sqrt{\frac{4-y^2}{2}}$
  • C
    $\frac{\sqrt{4+y^2}}{2}$
  • D
    $\frac{\sqrt{4-y^2}}{2}$

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$\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$ को हल करें।

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यदि $\cos^{-1} x = y$ है, तो $\dots \dots \dots$

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