The value of cot ^{-1}left(frac{sqrt{1+tan ^2 | vedclass

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(A) Let $f(x) = \cot ^{-1}\left(\frac{\sqrt{1+	an ^2 x}-1}{	an x}ight)$. Since $2$ is in the second quadrant,$\sec 2 < 0$,so $\sqrt{1+	an^2 2} =

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$\pi-\frac{5}{4}$

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(A) Let $f(x) = \cot ^{-1}\left(\frac{\sqrt{1+	an ^2 x}-1}{	an x}ight)$. Since $2$ is in the second quadrant,$\sec 2 For the second term,since $1/2$ is in the first quadrant,$\sec(1/2) > 0$,so $\sqrt{1+	an^2(1/2)} = \sec(1/2)$. Thus,the second term is $\cot ^{-1}\left(\frac{\sec(1/2) + 1}{	an(1/2)}ight) = \cot ^{-1}\left(\frac{1+\cos(1/2)}{\sin(1/2)}ight) = \cot ^{-1}\left(\cot(1/4)ight) = 1/4$. Subtracting the two,we get $(\pi - 1) - 1/4 = \pi - 5/4$.

The value of $\cot ^{-1}\left(\frac{\sqrt{1+\tan ^2(2)}-1}{\tan (2)}\right)-\cot ^{-1}\left(\frac{\sqrt{1+\tan ^2\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right)$ is equal to

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