$\tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}\left(\frac{x-y}{x+y}\right)$ નું સાદું રૂપ શું થાય?

  • A
    $0$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{2}$
  • D
    $\pi$

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Similar Questions

જો $\tan^{-1} \frac{1}{1+1(2)} + \tan^{-1} \frac{1}{1+2(3)} + \tan^{-1} \frac{1}{1+3(4)} + \dots + \tan^{-1} \frac{1}{1+n(n+1)} = \tan^{-1} \theta$ હોય,તો $\theta$ =

જો $\tan (x + y) = 33$ અને $x = \tan^{-1}(3)$ હોય,તો $y$ ની કિંમત શું થશે?

$\sin^{-1} \left( \frac{12}{13} \right) - \sin^{-1} \left( \frac{3}{5} \right)$ નું મૂલ્ય કેટલું થાય?

$x$ ના મૂલ્યોનો ગણ શોધો જેથી $\tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right)$ થાય.

જો $y = \sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) + \sec^{-1}\left(\frac{x^2 + 1}{x^2 - 1}\right)$,$|x| > 1$ હોય,તો $\frac{dy}{dx}$ ની કિંમત શોધો:

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