Show that $\tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right)=\frac{4-\sqrt{7}}{3}$ and justify why the other value $\frac{4+\sqrt{7}}{3}$ is ignored?
$$\begin{aligned} &\begin{aligned} \text { We have, }\quad\tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right) & =\frac{4-\sqrt{7}}{3} \\ \therefore\quad\text { LHS } & =\tan \left[\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)\right] \end{aligned} \end{aligned}$$
Let $$\frac{1}{2} \sin ^{-1} \frac{3}{4}=\theta \Rightarrow \sin ^{-1} \frac{3}{4}=2 \theta$$
$$\begin{array}{ll} \Rightarrow & \sin 2 \theta=\frac{3}{4} \Rightarrow \frac{2 \tan \theta}{1+\tan ^2 \theta}=\frac{3}{4} \\ \Rightarrow & 3+3 \tan ^2 \theta=8 \tan \theta \\ \Rightarrow & 3 \tan ^2 \theta-8 \tan \theta+3=0 \end{array}$$
$$\begin{aligned} &\begin{array}{r} \text { Let }\quad\tan \theta=y \\ \therefore\quad 3 y^2-8 y+3=0 \end{array} \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad y & =\frac{+8 \pm \sqrt{64-4 \times 3 \times 3}}{2 \times 3}=\frac{8 \pm \sqrt{28}}{6} \\ & =\frac{2[4 \pm \sqrt{7}]}{2 \cdot 3} \\ \Rightarrow \quad \tan \theta & =\frac{4 \pm \sqrt{7}}{3} \end{aligned}$$
$\Rightarrow \quad \theta=\tan ^{-1}\left[\frac{4 \pm \sqrt{7}}{3}\right]$
$\left\{\right.$ but $\frac{4+\sqrt{7}}{3}>\frac{1}{2} \cdot \frac{\pi}{2}$, since $\left.\max \left[\tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right)\right]=1\right\}$
$\therefore \quad$ LHS $=\tan \tan ^{-1}\left(\frac{4-\sqrt{7}}{3}\right)=\frac{4-\sqrt{7}}{3}=$ RHS
If $a_1, a_2, a_3, \ldots, a_n$ is an arithmetic progression with common difference $d$, then evaluate the following expression.
$$\begin{aligned} \tan \left[\tan ^{-1}\left(\frac{d}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{d}{1+a_2 a_3}\right)\right. & +\tan ^{-1}\left(\frac{d}{1+a_3 a_4}\right) \\ & \left.+\ldots+\tan ^{-1}\left(\frac{d}{1+a_{n-1} a_n}\right)\right] \end{aligned}$$
$$\begin{aligned} \text{We have,}\quad& a_1=a, a_2=a+d, a_3=a+2 d \\ \text{and}\quad & d=a_2-a_1=a_3-a_2=a_4-a_3=\ldots=a_n-a_{n-1} \end{aligned}$$
Given that, $$\tan \left[\tan ^{-1}\left(\frac{d}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{d}{1+a_2 a_3}\right)\right. \left.+\tan ^{-1}\left(\frac{d}{1+a_3 a_4}\right)+\ldots+\tan ^{-1}\left(\frac{d}{1+a_{n-1} \cdot a_n}\right)\right]$$
$$\begin{aligned} & =\tan \left[\tan ^{-1} \frac{a_2-a_1}{1+a_2 \cdot a_1}+\tan ^{-1} \frac{a_3-a_2}{1+a_3 \cdot a_2}+\ldots+\tan ^{-1} \frac{a_n-a_{n-1}}{1+a_n \cdot a_{n-1}}\right] \\ & =\tan \left[\left(\tan ^{-1} a_2-\tan ^{-1} a_1\right)+\left(\tan ^{-1} a_3-\tan ^{-1} a_2\right)+\ldots+\left(\tan ^{-1} a_n-\tan ^{-1} a_{n-1}\right)\right] \\ & =\tan \left[\tan ^{-1} a_n-\tan ^{-1} a_1\right] \\ & =\tan \left[\tan ^{-1} \frac{a_n-a_1}{1+a_n \cdot a_1}\right] \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right] \\ & =\frac{a_n-a_1}{1+a_n \cdot a_1} \quad\left[\because \tan \left(\tan ^{-1} x\right)=x\right] \end{aligned}$$
Which of the following is the principal value branch of $\cos ^{-1} x$ ?
Which of the following is the principal value branch of $\operatorname{cosec}^{-1} x$ ?
If $3 \tan ^{-1} x+\cot ^{-1} x=\pi$, then $x$ equals to