Q.If $|x|\le1$, then $2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ is equal to
(A) $4\tan^{-1}x$
(B) $0$
(C) $\frac{\pi}{2}$
(D) $\pi$

Reduce each angle to the principal range using the periodicity of $\tan$ and $\cot$ (period $\pi$) before applying the inverse function.

$\dfrac{13\pi}{6}=2\pi+\dfrac{\pi}{6}$, so $\tan\dfrac{13\pi}{6}=\tan\dfrac{\pi}{6}$, and $\tan^{-1}\left(\tan\dfrac{13\pi}{6}\right)=\dfrac{\pi}{6}$ (already in $(-\pi/2,\pi/2)$).

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Reduce the angle modulo $\pi$ (the period of $\tan$) until it lands inside $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

$\frac{31\pi}{6}=5\pi+\frac{\pi}{6}$. Since $\tan$ has period $\pi$, $\tan\frac{31\pi}{6}=\tan\frac{\pi}{6}$.

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Combine the inverse-cotangents, then apply $\cot$ and $\cos^{-1}$.

Use the tangent addition formula, then reject the root that makes the sum negative.

Expand the sum, convert each $\cot^{-1}$ of a negative argument using $\cot^{-1}(-x)=\pi-\cot^{-1}x$, then combine using the tangent addition formula.

$\displaystyle\sum_{i=0}^{2}\cot^{-1}\{-(i+1)\} = \cot^{-1}(-1)+\cot^{-1}(-2)+\cot^{-1}(-3)$.

Using $\cot^{-1}(-x)=\pi-\cot^{-1}x$ (principal range $(0,\pi)$):

$=[\pi-\cot^{-1}1]+[\pi-\cot^{-1}2]+[\pi-\cot^{-1}3] = 3\pi-[\cot^{-1}1+\cot^{-1}2+\cot^{-1}3]$.

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First combine the two inverse-trig terms into a single angle using the tangent addition formula, then simplify $\cot(2019\pi/2-\theta)$ using periodicity and a co-function identity.

Step 1 — combine the bracket. $\text{cosec}^{-1}\tfrac53$: if $\text{cosec}\,\theta=5/3$ then $\sin\theta=3/5,\cos\theta=4/5$, so $\tan\theta=3/4$, i.e. $\text{cosec}^{-1}\tfrac53=\tan^{-1}\tfrac34$.

So bracket $=\tan^{-1}\tfrac34+\tan^{-1}\tfrac23 = \tan^{-1}\!\left(\dfrac{3/4+2/3}{1-\tfrac34\cdot\tfrac23}\right)=\tan^{-1}\!\left(\dfrac{17/12}{1/2}\right)=\tan^{-1}\dfrac{17}{6}$.

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Use $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.

Convert to $\tan^{-1}$ and use the addition formula with the correct quadrant correction.

$\cot^{-1}\tfrac12=\tan^{-1}2$ and $\cot^{-1}\tfrac13=\tan^{-1}3$.

Since the product $2\cdot3=6>1$ and both are positive, …