We have, $\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}=\tan ^{-1} \frac{63}{16}$

Let $$\sin ^{-1} \frac{5}{13}=x$$

$$\begin{array}{lrl} \Rightarrow & \sin x & =\frac{5}{13} \\ \text { and } & \cos ^2 x & =1-\sin ^2 x \\ & \qquad=1-\frac{25}{169}=\frac{144}{169} \\ \Rightarrow & \cos x & =\sqrt{\frac{144}{169}}=\frac{12}{13} \\ \therefore & \tan x & =\frac{\sin x}{\cos x}=\frac{5 / 13}{12 / 13}=\frac{5}{12} \quad\text{.... (ii)}\\ \Rightarrow & \tan x & =5 / 12\quad\text{.... (iii)} \end{array}$$

$$\begin{aligned} \text{Again, let}\quad\cos ^{-1} \frac{3}{5} & =y \Rightarrow \cos y=\frac{3}{5} \\ \therefore\quad\sin y & =\sqrt{1-\cos ^2 y} \\ & =\sqrt{1-\left(\frac{3}{5}\right)^2}=\sqrt{1-\frac{9}{25}} \\ \sin y & =\sqrt{\frac{16}{25}}=\frac{4}{5} \\ \Rightarrow\quad\tan y & =\frac{\sin y}{\cos y}=\frac{4 / 5}{3 / 5}=\frac{4}{3}\quad\text{.... (iii)} \end{aligned}$$

$$\begin{aligned} &\text { We know that, }\\ &\begin{aligned} & \tan (x+y) =\frac{\tan x+\tan y}{1-\tan x \cdot \tan y} \\ \Rightarrow \quad & \tan (x+y) =\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \cdot \frac{4}{3}} \Rightarrow \tan (x+y)=\frac{\frac{15+48}{36}}{\frac{36-20}{36}} \\ \Rightarrow \quad & \tan (x+y) =\frac{63 / 36}{16 / 36} \\ \Rightarrow \quad & \tan (x+y) =\frac{63}{16} \\ \Rightarrow \quad & x+y =\tan ^{-1} \frac{63}{16} \\ \Rightarrow \quad & \tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{4}{3} =\tan ^{-1} \frac{63}{16} \end{aligned} \end{aligned}$$

Hence proved.

$$\begin{aligned} \text{We have,}\quad\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9} & =\sin ^{-1} \frac{1}{\sqrt{5}} \quad\text{.... (i)}\\ \text{Let}\quad\tan ^{-1} \frac{1}{4} & =x \\ \Rightarrow\quad\tan x & =\frac{1}{4} \\ \Rightarrow\quad\tan ^2 x & =\frac{1}{16} \\ \Rightarrow\quad\sec ^2 x-1 & =\frac{1}{16} \\ \Rightarrow\quad\sec ^2 x & =1+\frac{1}{16}=\frac{17}{16} \\ \Rightarrow\quad\frac{1}{\cos ^2 x} & =\frac{17}{16} \\ \Rightarrow\quad\cos ^2 x & =\frac{16}{17} \\ \Rightarrow\quad\cos ^x & =\frac{4}{\sqrt{17}} \\ \Rightarrow\quad\sin 2 x & =1-\cos ^2 x=1-\frac{16}{17}=\frac{1}{17} \\ \Rightarrow\quad\sin x & =\frac{1}{\sqrt{17}} \quad\text{.... (ii)} \end{aligned}$$

$$\begin{array}{lc} \text { Again, let } & \tan ^{-1} \frac{2}{9}=y \\ \Rightarrow & \tan y=\frac{2}{9} \Rightarrow \tan ^2 y=\frac{4}{81} \\ \Rightarrow & \sec ^2 y-1=\frac{4}{81} \\ \Rightarrow & \sec ^2 y=\frac{4}{81}+1=\frac{85}{81} \\ \Rightarrow & \cos ^2 y=\frac{81}{85} \Rightarrow \cos y=\frac{9}{\sqrt{85}} \\ \Rightarrow & \sin ^2 y=1-\cos ^2 y=1-\frac{81}{85}=\frac{4}{85} \\ \Rightarrow & \sin y=\frac{2}{\sqrt{85}}\quad\text{.... (iii)} \end{array}$$

$$\begin{aligned} \text { We know that, } \sin (x+y) & =\sin x \cdot \cos y+\cos x \cdot \sin y \\ & =\frac{1}{\sqrt{17}} \cdot \frac{9}{\sqrt{85}}+\frac{4}{\sqrt{17}} \cdot \frac{2}{\sqrt{85}} \\ & =\frac{17}{\sqrt{17} \cdot \sqrt{85}}=\frac{\sqrt{17}}{\sqrt{17} \cdot \sqrt{5}}=\frac{1}{\sqrt{5}} \\ \Rightarrow \quad(x+y) & =\sin ^{-1} \frac{1}{\sqrt{5}} \\ \Rightarrow \quad \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9} & =\sin ^{-1} \frac{1}{\sqrt{5}} \end{aligned}$$

Hence proved.

We have, $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}$

$$\begin{aligned} & =2 \cdot 2 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239} \quad \left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]\\ & =2 \cdot\left[\tan ^{-1} \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^2}\right]-\tan ^{-1} \frac{1}{239} \quad\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right] \\ & =2 \cdot\left[\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\frac{1}{25}}\right)\right]-\tan ^{-1} \frac{1}{239} \\ & =2 \cdot\left[\tan ^{-1}\left(\frac{2 / 5}{24 / 25}\right)\right]-\tan ^{-1} \frac{1}{239} \\ & =2 \tan ^{-1} \frac{5}{12}-\tan ^{-1} \frac{1}{239} \end{aligned}$$

$$ \begin{aligned} & =\tan ^{-1} \frac{2 \cdot \frac{5}{12}}{1-\left(\frac{5}{12}\right)^2}-\tan ^{-1} \frac{1}{239} \\ & =\tan ^{-1}\left(\frac{\frac{5}{6}}{1-\frac{25}{144}}\right)-\tan ^{-1} \frac{1}{239} \\ & =\tan ^{-1}\left(\frac{144 \times 5}{119 \times 6}\right)-\tan ^{-1} \frac{1}{239} \\ & =\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1} \frac{1}{239} \\ & =\tan ^{-1}\left(\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \cdot \frac{1}{239}}\right) \quad \left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]\\ & =\tan ^{-1}\left(\frac{120 \times 239-119}{119 \times 239+120}\right) \end{aligned}$$

$$\begin{aligned} & =\tan ^{-1}\left[\frac{28680-119}{28441+120}\right]=\tan ^{-1} \frac{28561}{28561} \\ & =\tan ^{-1}(1)=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { We have, }\quad\tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right) & =\frac{4-\sqrt{7}}{3} \\ \therefore\quad\text { LHS } & =\tan \left[\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)\right] \end{aligned} \end{aligned}$$

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

$$\begin{aligned} \text{We have,}\quad& a_1=a, a_2=a+d, a_3=a+2 d \\ \text{and}\quad & d=a_2-a_1=a_3-a_2=a_4-a_3=\ldots=a_n-a_{n-1} \end{aligned}$$

Given that, $$\tan \left[\tan ^{-1}\left(\frac{d}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{d}{1+a_2 a_3}\right)\right. \left.+\tan ^{-1}\left(\frac{d}{1+a_3 a_4}\right)+\ldots+\tan ^{-1}\left(\frac{d}{1+a_{n-1} \cdot a_n}\right)\right]$$

$$\begin{aligned} & =\tan \left[\tan ^{-1} \frac{a_2-a_1}{1+a_2 \cdot a_1}+\tan ^{-1} \frac{a_3-a_2}{1+a_3 \cdot a_2}+\ldots+\tan ^{-1} \frac{a_n-a_{n-1}}{1+a_n \cdot a_{n-1}}\right] \\ & =\tan \left[\left(\tan ^{-1} a_2-\tan ^{-1} a_1\right)+\left(\tan ^{-1} a_3-\tan ^{-1} a_2\right)+\ldots+\left(\tan ^{-1} a_n-\tan ^{-1} a_{n-1}\right)\right] \\ & =\tan \left[\tan ^{-1} a_n-\tan ^{-1} a_1\right] \\ & =\tan \left[\tan ^{-1} \frac{a_n-a_1}{1+a_n \cdot a_1}\right] \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right] \\ & =\frac{a_n-a_1}{1+a_n \cdot a_1} \quad\left[\because \tan \left(\tan ^{-1} x\right)=x\right] \end{aligned}$$