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### Answers (1)

Answer : $\frac{-3\pi}{4}$

Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$.

Given : $\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)+\tan ^{-1}(-\sqrt{3})+\tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)$

Explanation :

Let us first solve for $\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$

Let    $x=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$

\begin{aligned} &\tan x=\frac{-1}{\sqrt{3}} \\ &\tan x=-\tan \frac{\pi}{6} \end{aligned}                                                                                    $\left[\because \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$

$\tan x=\tan \left(\frac{-\pi}{6}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \tan (-\theta)=-\tan \theta]$

\begin{aligned} &x=\frac{-\pi}{6} \\ &\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} \end{aligned}                                                                   [From equation   (i) ]

The range of principal value branch of  $\tan ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.

$\because \; \; \; \; \tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Hence principal value of $\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6}$                                                      ... (ii)

Now,

Let us solve for   $\tan ^{-1}(-\sqrt{3})$

Let $y=\tan ^{-1}(-\sqrt{3})$                                                                                             ...(iii)

\begin{aligned} &\tan y=-\sqrt{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right] \\ &\tan y=-\tan \left(\frac{\pi}{3}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \tan (-\theta)=-\tan \theta] \end{aligned}

\begin{aligned} &\tan y=\tan \left(\frac{-\pi}{3}\right) \\ &y=\frac{-\pi}{3} \end{aligned}                                                                        [From equation (iii)]

$\tan ^{-1}(-\sqrt{3})=\frac{-\pi}{3}$

The Range of principal value branch of $\tan ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.

$\because \tan ^{-1}(-\sqrt{3})=\frac{-\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Hence principal value of $\tan ^{-1}(-\sqrt{3})=\frac{-\pi}{3}$                                                            ...(iv)

Now,

Let us solve for  $\tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)$

$\tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)=\tan ^{-1}(-1) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \sin \frac{\pi}{2}=1\right]$

\begin{aligned} &=\tan ^{-1}\left(\tan \left(\frac{-\pi}{4}\right)\right) \\ &=\frac{-\pi}{4} \\ \tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right) &=\frac{-\pi}{4} \end{aligned}

The Range of principal value branch of $\tan ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.

$\therefore \tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)=\frac{-\pi}{4} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Hence principal value $\tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)=\frac{-\pi}{4}$                                                          .....(v)

Now consider

$\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\tan ^{-1}(-\sqrt{3})+\tan ^{-1}\left(\sin \left(-\frac{\pi}{2}\right)\right)$

From equation (ii), (iv) and (v)

\begin{aligned} &=-\frac{\pi}{6}+\left(-\frac{\pi}{3}\right)+\left(\frac{-\pi}{4}\right) \\ &=-\frac{\pi}{6}-\frac{\pi}{3}-\frac{\pi}{4} \\ &=\frac{-2 \pi-4 \pi-3 \pi}{12} \\ &=\frac{-9 \pi}{12} \\ &=\frac{-3 \pi}{4} \end{aligned}

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