$\cot ^{-1}\left(\frac{\sqrt{1+\tan ^2(2)}-1}{\tan (2)}\right)-\cot ^{-1}\left(\frac{\sqrt{1+\tan ^2\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right)$ ની કિંમત શોધો.

  • A
    $\pi-\frac{5}{4}$
  • B
    $\pi-\frac{3}{2}$
  • C
    $\pi+\frac{3}{2}$
  • D
    $\pi+\frac{5}{2}$

Explore More

Similar Questions

જો $\pi \le x \le 2\pi $ હોય,તો ${\cos ^{ - 1}}(\cos x)$ કોના બરાબર થાય?

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ ની કિંમત શોધો.

જો $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$ હોય,તો $x$ ની કિંમત શોધો.

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$ ની કિંમત શોધો.

વિધાન $I:$ સમીકરણ $(\sin^{-1} x)^3 + (\cos^{-1} x)^3 - a\pi^3 = 0$ નો ઉકેલ તમામ $a \ge \frac{1}{32}$ માટે મળે છે.
વિધાન $II:$ કોઈપણ $x \in [-1, 1]$ માટે,$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ અને $0 \le (\sin^{-1} x - \frac{\pi}{4})^2 \le \frac{9\pi^2}{16}$ છે.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo