The value of ^-1(√1+^2 2-1 2) - ^-1(√1+^212+112) is equal to

Question

The value of $\cot^{-1}\left(\dfrac{\sqrt{1+	an^2 2}-1}{	an 2}ight) - \cot^{-1}\left(\dfrac{\sqrt{1+	an^2\frac12}+1}{	an\frac12}ight)$ is equal to

Accepted Answer

Since $\sqrt{1+\tan^2\theta}=|\sec\theta|$,

$$\cot^{-1}\left(\frac{|\sec2|-1}{\tan2}\right)-\cot^{-1}\left(\frac{\sec\frac12+1}{\tan\frac12}\right)=\cot^{-1}\left(\frac{-1-\cos2}{\sin2}\right)-\cot^{-1}\left(\frac{1+\cos\frac12}{\sin\frac12}\right).$$

Using half-angle identities, this simplifies to

$$\pi-\cot^{-1}(\cot1)-\cot^{-1}\left(\cot\frac14\right)=\pi-1-\frac14=\pi-\frac54.$$

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