Inverse Trigonometric Functions

Simplify the individual components

Let $\alpha=\sin ^{-1}\left(\frac{2}{\sqrt{13}}\right)$ and $\beta=\cos ^{-1}\left(\frac{3}{\sqrt{10}}\right)$.

• For $\alpha$ : Since $\sin \alpha=\frac{2}{\sqrt{13}}$, using Pythagoras' theorem (base $=\sqrt{(\sqrt{13})^2-2^2}=3$ ), we get:

$ \tan \alpha=\frac{2}{3} $

• For $\beta$ : Since $\cos \beta=\frac{3}{\sqrt{10}}$, using Pythagoras' theorem (perpendicular $= \sqrt{(\sqrt{10})^2-3^2}=1$ ), we get:

$ \tan \beta=\frac{1}{3} $

Rewrite the expression

The expression becomes: $\tan (2 \alpha-2 \beta)=\tan [2(\alpha-\beta)]$.

First, find $\tan (\alpha-\beta)$ using the formula $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$

$ \tan (\alpha-\beta)=\frac{\frac{2}{3}-\frac{1}{3}}{1+\left(\frac{2}{3} \cdot \frac{1}{3}\right)}=\frac{\frac{1}{3}}{1+\frac{2}{9}}=\frac{\frac{1}{3}}{\frac{11}{9}}=\frac{1}{3} \cdot \frac{9}{11}=\frac{3}{11} $

3. Calculate the final value

Now, use the double angle formula $\tan (2 \theta)=\frac{2 \tan \theta}{1-\tan ^2 \theta}$, where $\theta=(\alpha-\beta)$ :

$ \begin{gathered} \tan [2(\alpha-\beta)]=\frac{2\left(\frac{3}{11}\right)}{1-\left(\frac{3}{11}\right)^2} \\ \tan [2(\alpha-\beta)]=\frac{\frac{6}{11}}{1-\frac{9}{121}}=\frac{\frac{6}{11}}{\frac{112}{121}} \\ \tan [2(\alpha-\beta)]=\frac{6}{11} \cdot \frac{121}{112}=\frac{6 \cdot 11}{112}=\frac{66}{112}=\frac{33}{56} \end{gathered} $

Domain of $\sin ^{-1} \theta$ is $-1 \leq \theta \leq 1$

$f(x)=\sin ^{-1}\left(\frac{1}{x^2-2 x-2}\right)$ is defined if $-1 \leq \frac{1}{x^2-2 x-2} \leq 1$

s.t $\frac{1}{x^2-2 x-2} \geq-1$ and $\frac{1}{x^2-2 x-2} \leq 1$

so these two inequality and final solution will be intersection of solution of these two inequality.

solve $\frac{1}{x^2-2 x-2} \geq-1$

$ \begin{aligned} & \frac{1}{x^2-2 x-2}+1 \geq 0 \\ & \frac{1+x^2-2 x-2}{x^2-2 x-2} \geq 0 \\ & \frac{\left(x^2-2 x-1\right)}{x^2-2 x-2} \geq 0 \end{aligned} $

roots of $x^2-2 x-1$ is $x=\frac{2 \pm \sqrt{4+4}}{2}=1 \pm \sqrt{2}$

roots of $x^2-2 x-2$ is $x=\frac{2 \pm \sqrt{4+8}}{2}=1 \pm \sqrt{3}$

$ \begin{aligned} & \text { so, } x \in(-\infty, 1-\sqrt{3}) \cup[1-\sqrt{2}, 1+\sqrt{2}] \cup(1+\sqrt{3}, \infty)-S_1 \\ & \text { solve } \frac{1}{x^2-2 x-2} \leq 1 \\ & \frac{1}{x^2-2 x-2}-1 \leq 0 \\ & \frac{1-x^2+2 x+2}{x^2-2 x-2} \leq 0 \\ & \frac{3-x^2+2 x}{x^2-2 x-2} \leq 0 \\ & \frac{x^2-2 x-3}{x^2-2 x-2} \geq 0 \text { multiply by }-1 \text { sign of inequality change } \\ & \text { roots of } x^2-2 x-3 \text { is } x=\frac{2 \pm \sqrt{4+12}}{2}=3,-1 \\ & \text { roots of } x^2-2 x-2 \text { is } x=1 \pm \sqrt{3} \end{aligned} $

$ x \in(-\infty,-1] \cup(1-\sqrt{3}, 1+\sqrt{3}) \cup[3, \infty)-S_2 $

taking intersection of $S_1 \& S_2$ for final solution.

so domain of $f(x)$ is $x \in(-\infty,-1] \cup[1-\sqrt{2}, 1+\sqrt{2}] \cup[3, \infty)$

compare this with given domain $x \in(-\infty, \alpha] \cup[\beta, \gamma] \cup[\delta, \infty)$

$ \begin{aligned} & \text { so, } \alpha=-1, \beta=1-\sqrt{2}, \gamma=1+\sqrt{2}, \delta=3 \\ & \alpha+\beta+\gamma+\delta=-1+1-\sqrt{2}+1+\sqrt{2}+3=4 \end{aligned} $

$ \begin{aligned} &\tan ^{-1} 4 x+\tan ^{-1} 6 x=\frac{\pi}{6}\\ &\begin{aligned} & \tan ^{-1}\left(\frac{10 x}{1-24 x^2}\right)=\frac{\pi}{6} \\ & 24 x^2+10 \sqrt{3} x-1=0 \\ & x=\frac{-5 \sqrt{3} \pm 3 \sqrt{14}}{24} \end{aligned} \end{aligned} $

$ \text { (1) } \frac{2 x-5}{11-3 x}-1 \leq 0 $

$ \begin{aligned} & \Rightarrow \frac{2 x-5-11+3 x}{11-3 x} \leq 0 \\ & \Rightarrow \frac{5 x-16}{11-3 x} \leq 0 \\ & \Rightarrow x \in\left(-\infty, \frac{16}{5}\right] \cup\left(\frac{11}{3}, \infty\right) \ldots \text { (i) } \end{aligned} $

$ \text { } \begin{aligned} (2) & \frac{2 x-5}{11-3 x} \geq-1 \\ & \Rightarrow \frac{2 x-5}{11-3 x}+1 \geq 0 \\ & \Rightarrow \frac{6-x}{11-3 x} \geq 0 \\ & \Rightarrow \frac{x-6}{3 x-11} \geq 0 \\ & \Rightarrow x \in\left(-\infty, \frac{11}{3}\right) \cup[6, \infty) \ldots \text { (ii) } \end{aligned} $

$ \text { } \begin{array}{r}(3) 2 x^2-3 x+1 \leq 1 \\ \Rightarrow \quad x(2 x-3) \leq 0 \\ x \in\left[0, \frac{3}{2}\right] \end{array} $

$ \text { } \begin{gathered}(4) 2 x^2-3 x+1 \geq-1 \\ \Rightarrow 2 x^2-3 x+2 \geq 0 \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x \in R \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{gathered} $

$ \begin{aligned} &\text { taking intersection of (i), (ii), (iii), (iv) }\\ &\begin{aligned} x & \in\left[0, \frac{3}{2}\right] \\ \therefore \quad & \alpha+2 \beta \\ & =3 \end{aligned} \end{aligned} $

$\cot ^{-1}\left(\frac{|\sec 2|-1}{\tan 2}\right)-\cot ^{-1}\left(\frac{\left|\sec \frac{1}{2}\right|+1}{\tan \frac{1}{2}}\right)$

$\begin{aligned} & =\cot ^{-1}\left(\frac{-1-\cos 2}{\sin 2}\right)-\cot ^{-1}\left(\frac{1+\cos \frac{1}{2}}{\sin \frac{1}{2}}\right) \\ & =\pi-\cot ^{-1}(\cot 1)-\cot ^{-1}\left(\cot \frac{1}{4}\right) \\ & =\pi-1-\frac{1}{4}=\pi-\frac{5}{4} \end{aligned}$

$\begin{aligned} & \cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots \\ & T_r=\cot ^{-1}\left(\frac{4 r^2+3}{4}\right) \\ & T_r=\tan ^{-1}\left(\frac{1}{\left(\frac{3}{4}+r^2\right)}\right) \end{aligned}$

$\begin{aligned} & T_r=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+r^2-1 / 4}\right) \\ & T_r=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right) \\ & T_r=\tan ^{-1}\left(r+\frac{1}{2}\right)-\tan ^{-1}\left(r-\frac{1}{2}\right) \\ & T_1=\tan ^{-1}\left(\frac{3}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}$

$\begin{aligned} & T_2=\tan ^{-1}\left(\frac{5}{2}\right)-\tan ^{-1}\left(\frac{3}{2}\right) \\ & \vdots \qquad\qquad \vdots \qquad \qquad\qquad \vdots \\ & T_n=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\ & \Sigma T_r=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\ & \Sigma T_r=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}$

$\begin{aligned} & \sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right),-\frac{1}{2}< x<\frac{1}{\sqrt{2}} \\ & \text { Let } x=\cos \theta, \theta \in\left(\frac{\pi}{4}, \frac{2 \pi}{3}\right) \\ & \Rightarrow \sqrt{1-x^2}=\sin \theta \text { as } \sin \theta>0 \end{aligned}$

$\begin{aligned} \sin ^{-1}\left(\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta\right)=\sin ^{-1}( & \left.\sin \left(\frac{\pi}{3}+\theta\right)\right) \\ & \frac{\pi}{3}+\theta \in\left(\frac{7 \pi}{2}, \pi\right)\end{aligned}$

$\begin{aligned} & =\sin ^{-1}\left(\sin \left(\pi-\left(\frac{\pi}{3}+\theta\right)\right)\right) \\ & =\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}-\theta\right)\right) \\ & =\frac{2 \pi}{3}-\theta \\ & =\frac{2 \pi}{3}-\cos ^{-1} x \\ & =\frac{2 \pi}{3}-\left(\frac{\pi}{2}-\sin ^{-1} x\right) \\ & =\frac{\pi}{6}+\sin ^{-1} x \end{aligned}$

$\begin{aligned} & 2[\mathrm{x}]+1 \leq-1 \text { or } 2[\mathrm{x}]+1 \geq 1 \\ & \Rightarrow[\mathrm{x}] \leq-1 \cup[\mathrm{x}] \geq 0 \\ & \Rightarrow \mathrm{x} \in(-\infty, 0) \cup \mathrm{x} \in[0, \infty) \\ & \Rightarrow \mathrm{x} \in(-\infty, \infty) \end{aligned}$

$\begin{aligned} & \cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right) \\ & \cos \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{33}{56}\right) \\ & \cos \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{5}{12}}{1+\frac{3}{4} \cdot \frac{5}{12}}\right)+\tan ^{-1} \frac{33}{56}\right) \\ & \cos \left(\tan ^{-1} \frac{56}{33}+\cot ^{-1} \frac{56}{33}\right) \\ & \cos \left(\frac{\pi}{2}\right)=0 \end{aligned}$

$\begin{aligned} & \Rightarrow \cot ^{-1}\left(\frac{\alpha \beta+1}{\alpha-\beta}\right)+\cot ^{-1}\left(\frac{\beta \gamma+1}{\beta-\gamma}\right)+\cot ^{-1}\left(\frac{\alpha \gamma+1}{\gamma-\alpha}\right) \\ & \Rightarrow \tan ^{-1}\left(\frac{\alpha-\beta}{1+\alpha \beta}\right)+\tan ^{-1}\left(\frac{\beta-\gamma}{1+\beta \gamma}\right)+\pi+\tan ^{-1}\left(\frac{\gamma-\alpha}{1+\gamma \alpha}\right) \\ & \Rightarrow\left(\tan ^{-1} \alpha-\tan ^{-1} \beta\right)+\left(\tan ^{-1} \beta-\tan ^{-1} \gamma\right)+\left(\pi+\tan ^{-1} \gamma-\tan ^{-1} \alpha\right) \\ & \Rightarrow \pi \end{aligned}$

$\begin{aligned} & \frac{\pi}{2} \leq x \leq \frac{3 \pi}{4} \\ & \cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{12} \sin x\right) \\ & \cos ^{-1}(\cos x \cos \alpha+\sin x \sin \alpha) \\ & \cos ^{-1}(\cos (x-\alpha)) \\ & \Rightarrow x-\alpha \text { because } x-\alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ & \Rightarrow x-\tan ^{-1} \frac{5}{12} \end{aligned}$

Let $ f(x) = 16\left[\left(\sec^{-1} x\right)^2 + \left(\operatorname{cosec}^{-1} x\right)^2\right] $.

We can express $ f(x) $ as:

$ f(x) = 16\left[\left(\sec^{-1} x + \operatorname{cosec}^{-1} x\right)^2 - 2\left(\sec^{-1} x\right)\left(\frac{\pi}{2} - \sec^{-1} x\right)\right] $

This simplifies to:

$ f(x) = 16\left[\frac{\pi^2}{4} - \pi \sec^{-1} x + 2 \left(\sec^{-1} x\right)^2\right], \quad \text{where } \sec^{-1} x \in [0, \pi] - \left\{\frac{\pi}{2}\right\} $

Further simplification gives:

$ f(x) = 16\left[2\left(\sec^{-1} x - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{4} - \frac{\pi^2}{8}\right] $

For the maximum value when $ \sec^{-1} x = \pi $:

$ \max = 16\left[2\pi^2 - \pi^2 + \frac{\pi^2}{4}\right] = 20\pi^2 $

For the minimum value when $ \sec^{-1} x = \frac{\pi}{4} $:

$ \min = 16\left[\frac{2 \times \pi^2}{16} - \frac{\pi^2}{4} + \frac{\pi^2}{4}\right] = 2\pi^2 $

Therefore, the sum of the maximum and minimum values is:

$ \text{Sum} = 22\pi^2 $

$\begin{aligned} & \cos ^{-1} x-\frac{\pi}{2}+\cos ^{-1} y=\alpha \\ & \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\ & \because \quad \alpha \in\left[\frac{-\pi}{2}, \pi\right] \\ & \text { then } \frac{\pi}{2}+\alpha \in\left(0, \frac{3 \pi}{2}\right) \\ & \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\frac{\pi}{2}+\alpha \\ & x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha \\ & x y+\sin \alpha=\sqrt{1-x^2} \sqrt{1-y^2} \\ & x^2 y^2+\sin ^2 \alpha+2 x y \sin \alpha=1-x^2-y^2+x^2 y^2 \\ & \underbrace{x^2+y^2+2 x y \sin \alpha}_E=\cos ^2 \alpha \end{aligned}$

Now, minimum value of $E$ is 0.

$\begin{aligned} & \sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _e\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right) \\ & -1 \leq \frac{3 x-22}{2 x-19} \leq 1 \\ & \frac{3 x-22}{2 x-19}+1 \geq 0 \text { and } \frac{3 x-22}{2 x-19}-1 \leq 0 \\ & \frac{3 x-22+2 x-19}{2 x-19} \geq 0 \text { and } \frac{3 x-22-2 x+19}{2 x-19} \leq 0 \\ & \Rightarrow \frac{5 x-41}{2 x-19} \geq 0 \text { and } \frac{x-3}{2 x-19} \leq 0 \\ & x \in\left(-\infty, \frac{41}{5}\right] \cup\left(\frac{19}{2}, \infty\right) \text { and } x \in\left[3, \frac{19}{2}\right) \end{aligned}$

$\begin{aligned} & \Rightarrow \quad x \in\left[3, \frac{41}{5}\right] \quad \text{.... (1)}\\ & \text { and, } \frac{3 x^2-8 x+5}{x^2-3 x-10}>0 \\ & \frac{(3 x-5)(x-1)}{(x-5)(x-2)}>0 \\ & \Rightarrow \quad x \in(-\infty,-2) \cup\left[1, \frac{5}{3}\right] \cup(5, \infty) \ldots \end{aligned}$

Taking intersection of individual domains

$x \in\left(5, \frac{41}{5}\right]$

$\begin{aligned} & \Rightarrow \quad \alpha=5 \text { and } \beta=\frac{41}{5} \\ & \Rightarrow 3 \alpha+10 \beta=15+82 \\ & =97 \end{aligned}$

$\therefore \quad$ Option (4) is correct

$\begin{aligned} & a=\sin ^{-1}(\sin 5)=5-2 \pi \\ & \text { and } b=\cos ^{-1}(\cos 5)=2 \pi-5 \\ & \therefore a^2+b^2=(5-2 \pi)^2+(2 \pi-5)^2 \\ & =8 \pi^2-40 \pi+50 \end{aligned}$

Let $\sin ^{-1} \alpha=A, \sin ^{-1} \beta=B, \sin ^{-1} \gamma=C$

$\begin{aligned} & \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\ & (\alpha+\beta)^2-\gamma^2=3 \alpha \beta \\ & \alpha^2+\beta^2-\gamma^2=\alpha \beta \\ & \frac{\alpha^2+\beta^2-\gamma^2}{2 \alpha \beta}=\frac{1}{2} \\ & \Rightarrow \cos \mathrm{C}=\frac{1}{2} \\ & \sin \mathrm{C}=\gamma \\ & \cos \mathrm{C}=\sqrt{1-\gamma^2}=\frac{1}{2} \\ & \gamma=\frac{\sqrt{3}}{2} \end{aligned}$

Assume $\sin ^{-1} x=\theta$

$\begin{aligned} & \cos (2 \theta)=\frac{1}{9} \\ & \sin \theta= \pm \frac{2}{3} \end{aligned}$

as $\mathrm{m}$ and $\mathrm{n}$ are co-prime natural numbers,

$\mathrm{x}=\frac{2}{3}$

i.e. $m=2, n=3$

So, the quadratic equation becomes $2 x^2-3 x+1=0$ whose roots are $\alpha=1, \beta=\frac{1}{2}$

$\left(1, \frac{1}{2}\right)$ lies on $5 x+8 y=9$

$\begin{aligned} & \tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4} ; x>0 \\ & \Rightarrow \tan ^{-1} 2 x=\frac{\pi}{4}-\tan ^{-1} x \end{aligned}$

Taking tan both sides

$\begin{aligned} & \Rightarrow 2 \mathrm{x}=\frac{1-\mathrm{x}}{1+\mathrm{x}} \\ & \Rightarrow 2 \mathrm{x}^2+3 \mathrm{x}-1=0 \\ & \mathrm{x}=\frac{-3 \pm \sqrt{9+8}}{8}=\frac{-3 \pm \sqrt{17}}{8} \end{aligned}$

Only possible $x=\frac{-3+\sqrt{17}}{8}$

To find the domain of the function, we need to consider the individual functions and their respective domains. We have:

1. $f_1(x) = \ln(4x^2 + 11x + 6)$
2. $f_2(x) = \sin^{-1}(4x + 3)$
3. $f_3(x) = \cos^{-1}\left(\frac{10x + 6}{3}\right)$

1. For $f_1(x)$:

$ 4x^2 + 11x + 6 > 0 $

Factoring the quadratic expression:

$ (4x + 3)(x + 2) > 0 $

From this inequality, we have:

$ x \in (-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right) $

1. For $f_2(x)$:

$ -1 \le 4x + 3 \le 1 $

From these inequalities, we get:

$ x \in \left[-1, -\frac{1}{2}\right] $

1. For $f_3(x)$:

$ -1 \le \frac{10x + 6}{3} \le 1 $

From these inequalities, we get:

$ x \in \left[-\frac{9}{10}, -\frac{3}{10}\right] $

Now, we need to find the intersection of the domains of the three functions:

$ \left(-\infty, -2\right) \cup \left(-\frac{3}{4}, \infty\right) \cap \left[-1, -\frac{1}{2}\right] \cap \left[-\frac{9}{10}, -\frac{3}{10}\right] $

To find the intersection, let's analyze the intervals:

• The interval $(-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right)$ contains all $x$ values less than $-2$ and greater than $-\frac{3}{4}$.
• The interval $\left[-1, -\frac{1}{2}\right]$ contains all $x$ values between $-1$ and $-\frac{1}{2}$.
• The interval $\left[-\frac{9}{10}, -\frac{3}{10}\right]$ contains all $x$ values between $-\frac{9}{10}$ and $-\frac{3}{10}$.

Looking at the intervals, we can see that the intersection is:

$ x \in \left(-\frac{3}{4}, -\frac{1}{2}\right] $

Thus, the domain of the function is $(\alpha, \beta] = \left(-\frac{3}{4}, -\frac{1}{2}\right]$. Now, we need to find the value of $36|\alpha + \beta|$:

$ 36\left|-\frac{3}{4} - \frac{1}{2}\right| = 36\left|-\frac{5}{4}\right| =45 $

${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$

$= {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{16 + 8\sqrt 3 } \over {12 + 6\sqrt 3 }}} \right)^{{1 \over 2}}}$

$ = {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {\sqrt 3 (\sqrt 3 + 1)}}} \right) + {\sec ^{ - 1}}{\left( {{{4({1^2} + {{(\sqrt 3 )}^2} + 2\,.\,1\,.\,\sqrt 3 } \over {{3^2}{{(\sqrt 3 )}^2} + 2\,.\,3\,.\,\sqrt 3 }}} \right)^{{1 \over 2}}}$

$ = {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{4{{(\sqrt 3 + 1)}^2}} \over {{{(3 + \sqrt 3 )}^2}}}} \right)^{{1 \over 2}}}$

$ = {\pi \over 6} + {\sec ^{ - 1}}\left( {{{2(\sqrt 3 + 1)} \over {\sqrt 3 (\sqrt 3 + 1)}}} \right)$

$ = {\pi \over 6} + {\sec ^{ - 1}}\left( {{2 \over {\sqrt 3 }}} \right)$

$ = {\pi \over 6} + {\cos ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right)$

$ = {\pi \over 6} + {\pi \over 6}$

$ = {\pi \over 3}$

$f(x) = {\tan ^{ - 1}}(\sin x - \cos x),\,\,\,\,\,[0,\pi ]$

Let $g(x) = \sin x - \cos x$

$ = \sqrt 2 \sin \left( {x - {\pi \over 4}} \right)$ and $x - {\pi \over 4} \in \left[ {{{ - \pi } \over 4},\,{{3\pi } \over 4}} \right]$

$\therefore$ $g(x) \in \left[ { - 1,\,\sqrt 2 } \right]$

and ${\tan ^{ - 1}}x$ is an increasing function

$\therefore$ $f(x) \in \left[ {{{\tan }^{ - 1}}( - 1),\,{{\tan }^{ - 1}}\sqrt 2 } \right]$

$ \in \left[ { - {\pi \over 4},\,{{\tan }^{ - 1}}\sqrt 2 } \right]$

$\therefore$ Sum of ${f_{\max }}$ and ${f_{\min }} = {\tan ^{ - 1}}\sqrt 2 - {\pi \over 4}$

$ = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) - {\pi \over 4}$

$ - 1 \le {{{x^2} - 4x + 2} \over {{x^2} + 3}} \le 1$

$ \Rightarrow - {x^2} - 3 \le {x^2} - 4x + 2 \le {x^2} + 3$

$ \Rightarrow 2{x^2} - 4x + 5 \ge 0$ & $ - 4x \le 1$

$x \in R$ & $x \ge - {1 \over 4}$

So domain is $\left[ { - {1 \over 4},\infty } \right)$

${\cos ^{ - 1}}x - 2{\sin ^{ - 1}}x = {\cos ^{ - 1}}2x$

For Domain : $x \in \left[ {{{ - 1} \over 2},{1 \over 2}} \right]$

${\cos ^{ - 1}}x - 2\left( {{\pi \over 2} - {{\cos }^{ - 1}}x} \right) = {\cos ^{ - 1}}(2x)$

$ \Rightarrow {\cos ^{ - 1}}x + 2{\cos ^{ - 1}}x = \pi + {\cos ^{ - 1}}2x$

$ \Rightarrow \cos (3{\cos ^{ - 1}}x) = - \cos ({\cos ^{ - 1}}2x)$

$ \Rightarrow 4{x^3} = x$

$ \Rightarrow x = 3,\, \pm \,{1 \over 2}$

$ - 1 \le 2{x^2} - 3 < 2$

or $2 \le 2{x^2} < 5$

or $1 \le {x^2} < {5 \over 2}$

$x \in \left( { - \sqrt {{5 \over 2}} , - 1} \right] \cup \left[ {1,\sqrt {{5 \over 2}} } \right)$

${\log _{{1 \over 2}}}({x^2} - 5x + 5) > 0$

$0 < {x^2} - 5x + 5 < 1$

${x^2} - 5x + 5 > 0$ & ${x^2} - 5x + 4 < 0$

$x \in \left( { - \infty ,{{5 - \sqrt 5 } \over 2}} \right) \cup \left( {{{5 + \sqrt 5 } \over 2},\infty } \right)$

& $x \in ( - \infty ,1) \cup (4,\infty )$

Taking intersection

$x \in \left( {1,{{5 - \sqrt 5 } \over 2}} \right)$

Let ${{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta } = k \Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = k(\alpha + \beta )$

$ \Rightarrow \alpha + \beta = {\pi \over {2k}}$

Now, ${{2\pi \,\alpha } \over {\alpha + \beta }} = {{2\pi \,\alpha } \over {{\pi \over {2k}}}} = 4k\alpha = 4{\sin ^{ - 1}}x$

Here $\sin \left( {{{2\pi \,\alpha } \over {\alpha + \beta }}} \right) = \sin (4{\sin ^{ - 1}}x)$

Let ${\sin ^{ - 1}}x = \theta $

$\because$ $x \in \left( {0,{1 \over {\sqrt 2 }}} \right) \Rightarrow \theta \in \left( {0,{\pi \over 4}} \right)$

$ \Rightarrow x = \sin \theta $

$ \Rightarrow \cos \theta = \sqrt {1 - {x^2}} $

$ \Rightarrow \sin 2\theta = 2x\,.\,\sqrt {1 - {x^2}} $

$ \Rightarrow \cos 2\theta = \sqrt {1 - 4{x^2}(1 - {x^2})} = \sqrt {{{(2{x^2} - 1)}^2}} = 1 - 2{x^2}$

$\because$ $\left( {\cos 2\theta > 0\,\mathrm{as}\,2\theta \in \left( {0,{\pi \over 2}} \right)} \right)$

$ \Rightarrow \sin 4\theta = 2\,.\,2x\sqrt {1 - {x^2}} (1 - 2{x^2})$

$ = 4x\sqrt {1 - {x^2}} (1 - 2{x^2})$

$\tan \left( {2{{\tan }^{ - 1}}{1 \over 5} + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2} + 2{{\tan }^{ - 1}}{1 \over 8}} \right)$

$ = \tan \left( {2{{\tan }^{ - 1}}\left( {{{{1 \over 5} + {1 \over 8}} \over {1 - {1 \over 5}\,.\,{1 \over 8}}}} \right) + {{\sec }^{ - 1}}{{\sqrt 5 } \over 2}} \right)$

$ = \tan \left[ {2{{\tan }^{ - 1}}{1 \over 3} + {{\tan }^{ - 1}}{1 \over 2}} \right]$

$ = \tan \left[ {{{\tan }^{ - 1}}{{{2 \over 3}} \over {1 - {1 \over 9}}} + {{\tan }^{ - 1}}{1 \over 2}} \right]$

$ = \tan \left[ {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{1 \over 2}} \right]$

$ = \tan \left[ {{{\tan }^{ - 1}}{{{3 \over 4} + {1 \over 2}} \over {1 - {3 \over 8}}}} \right] = \tan \left[ {{{\tan }^{ - 1}}{{{5 \over 4}} \over {{5 \over 8}}}} \right]$

$ = \tan [{\tan ^{ - 1}}2] = 2$

$f(x) = {\sin ^{ - 1}}(2x) + \sin 2x + {\cos ^{ - 1}}(2x) + \cos 2x$

$ = {\sin ^{ - 1}}(2x) + {\cos ^{ - 1}}(2x) + \sin 2x + \cos 2x$

$ = {\pi \over 2} + \sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2x + {1 \over {\sqrt 2 }}\cos 2x} \right)$

$ = {\pi \over 2} + \sqrt 2 \left( {\cos {\pi \over 4}\sin 2x + \sin {\pi \over 4}\cos 2x} \right)$

$ = {\pi \over 2} + \sqrt 2 \,.\,\sin \left( {2x + {\pi \over 4}} \right)$

f(x) is maximum when $\sin \left( {2x + {\pi \over 4}} \right)$ is maximum means $x = {\pi \over 8}$ or $\sin \left( {2 \times {\pi \over 8} + {\pi \over 4}} \right) = \sin {\pi \over 2} = 1$

$\therefore$ ${\left[ {f(x)} \right]_{\max }} = {\pi \over 2} + \sqrt 2 \,.\,1 = {\pi \over 2} + \sqrt 2 = M$

f(x) is minimum when $\sin \left( {2x + {\pi \over 4}} \right)$ is minimum means $x = 0$ or $\sin \left( {2 \times 0 + {\pi \over 4}} \right) = {1 \over {\sqrt 2 }}$

$\therefore$ ${\left[ {f(x)} \right]_{\min }} = {\pi \over 2} + \sqrt 2 \,.\,{1 \over {\sqrt 2 }} = {\pi \over 2} + 1 = m$

$\therefore$ $m + M = {\pi \over 2} + \sqrt 2 + {\pi \over 2} + 1 = \pi + \sqrt 2 + 1$

Given,

$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$

We know, $2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1)$

$\therefore$ $2{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) = {\cos ^1}\left( {2 \times {1 \over 5} - 1} \right) = {\cos ^{ - 1}}\left( { - {3 \over 5}} \right)$

$\therefore$ $\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( { - {3 \over 5}} \right)} \right.} \right)$

$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {\pi - {{\cos }^{ - 1}}\left( { {3 \over 5}} \right)} \right.} \right)$

$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right.} \right)$

$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right)} \right.} \right)$

$ = \tan \left( {{{5\pi } \over {16}} \times {4 \over 5}} \right)$

$ = \tan \left( {{\pi \over 4}} \right)$

$ = 1$

$\therefore$ $\alpha = 1$

Also given,

$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$

$ = \cos \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$

$ = \cos \left( {2{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$

$ = \cos \left( {{{\cos }^{ - 1}}\left( {2\left. {{{\left( {{3 \over 5}} \right)}^2} - 1} \right)} \right.} \right)$

$ = \cos \left( {{{\cos }^{-1}}\left( {{{18} \over {25}} - 1} \right.} \right)$

$ = {{18} \over {25}} - 1$

$ = {{18 - 25} \over {25}}$

$ = - {7 \over {25}}$

$\therefore$ $\beta = - {7 \over {25}}$

$\therefore$ The quadratic equation with roots $\alpha$ and $\beta$ is

${x^2} - (\alpha + \beta )x + \alpha \beta = 0$

$ \Rightarrow {x^2} - \left( {1 - {7 \over {25}}} \right)x + 1 \times \left( { - {7 \over {25}}} \right) = 0$

$ \Rightarrow 25{x^2} - 18x - 7 = 0$

$ - 1 \le {{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi } \le 1$

$ \Rightarrow - {\pi \over 2} \le {\sin ^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right) \le {\pi \over 2}$

$ \Rightarrow - 1 \le {1 \over {4{x^2} - 1}} \le 1$

$\therefore$ ${1 \over {4{x^2} - 1}} + 1 \ge 0$

$ \Rightarrow {{1 + 4{x^2} - 1} \over {4{x^2} - 1}} \ge 0$

$ \Rightarrow {{4{x^2}} \over {4{x^2} - 1}} \ge 0$

$ \Rightarrow $ ${{4{x^2}} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0$ ...... (1)

$\therefore$ $x \in \left( { - \alpha , - {1 \over 2}} \right) \cup \{ 0\} \cup \left( {{1 \over 2},\alpha } \right)$ .....(2)

And ${1 \over {4{x^2} - 1}} - 1 \le 0$

$ \Rightarrow {{1 - 4{x^2} + 1} \over {4{x^2} - 1}} \le 0$

$ \Rightarrow {{2 - 4{x^2}} \over {4{x^2} - 1}} \le 0$

$ \Rightarrow {{2{x^2} - 1} \over {4{x^2} - 1}} \ge 0$

$ \Rightarrow $ ${{\left( {\sqrt 2 x + 1} \right)\left( {\sqrt 2 x - 1} \right)} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0$ ...... (3)

$x \in \left( { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left( { - {1 \over 2},{1 \over 2}} \right) \cup \left( {{1 \over {\sqrt 2 }},\alpha } \right)$ .....(4)

From (3) and (4), we get

$\therefore$ $x \in \left[ { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left[ {{1 \over {\sqrt 2 }},\alpha } \right) \cup \{ 0\} $

$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(n + 1) - n} \over {1 + (n + 1)n}}} \right)} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{50} {({{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}n} } \right)$

$ = \cot ({\tan ^{ - 1}}51 - {\tan ^{ - 1}}1)$

$ = \cot \left( {{{\tan }^{ - 1}}\left( {{{51 - 1} \over {1 + 51}}} \right)} \right)$

$ = \cot \left( {{{\cot }^{ - 1}}\left( {{{52} \over {50}}} \right)} \right)$

$ = {{26} \over {25}}$

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

$ = {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4}$

$ = {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}$

${\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right)$

$ = {\cos ^{ - 1}}\left( {{3 \over {10}}\,.\,{3 \over 5} + {2 \over 5}\,.\,{4 \over 5}} \right)$

$ = {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}$

${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin {\pi \over 4}}}} \right)$

$ = {\tan ^{ - 1}}\left( {{{{1 \over {\sqrt 2 }} - 1} \over {{1 \over {\sqrt 2 }}}}} \right)$

$ = {\tan ^{ - 1}}(1 - \sqrt 2 ) = - {\tan ^{ - 1}}(\sqrt 2 - 1)$

$ = - {\pi \over 8}$

Given,

$x\, * \,y = {x^2} + {y^3}$

$\therefore$ $x\, * \,1 = {x^2} + {1^3} = {x^2} + 1$

Now, $(x\, * \,1)\, * \,1 = ({x^2} + 1)\, * \,1$

$ \Rightarrow (x\, * \,1)\, * \,1 = {({x^2} + 1)^2} + {1^3}$

$ \Rightarrow (x\, * \,1)\, * \,1 = {x^4} + 1 + 2{x^2} + 1$

Also, $x\, * \,(1\, * \,1)$

$ = x\, * \,({1^2} + {1^3})$

$ = x\, * \,2$

$ = {x^2} + {2^3}$

$ = {x^2} + 8$

Given that,

$(x\, * \,1)\, * \,1 = x\, * \,(1\, * \,1)$

$\therefore$ ${x^4} + 1 + 2{x^2} + 1 = {x^2} + 8$

$ \Rightarrow {x^4} + {x^2} - 6 = 0$

$ \Rightarrow {x^4} + 3{x^2} - 2{x^2} - 6 = 0$

$ \Rightarrow {x^2}({x^2} + 3) - 2({x^3} + 3) = 0$

$ \Rightarrow ({x^2} + 3)({x^2} - 2) = 0$

$ \Rightarrow {x^2} = 2,\, - 3$

[${x^2} = -3$ not possible as square of anything should be always positive]

$\therefore$ ${x^2} = 2$

$\therefore$ Now,

$2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$

$ = 2{\sin ^{ - 1}}\left( {{{{2^2} + 2 - 2} \over {{2^2} + 2 + 2}}} \right)$

$ = 2{\sin ^{ - 1}}\left( {{4 \over 8}} \right)$

$ = 2{\sin ^{ - 1}}\left( {{1 \over 2}} \right)$

$ = 2 \times {\pi \over 6}$

$ = {\pi \over 3}$

${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3}$

Let $f(t) = {t^3} + {\left( {{\pi \over 2} - t} \right)^3}$

Where $t = {\tan ^{ - 1}}x$ ; $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$

$ = {t^3} + {\left( {{\pi \over 2}} \right)^3} - {{3{\pi ^2}t} \over 4} + {{3\pi } \over 2}{t^2} - {t^3}$

$f(t) = {{3\pi } \over 2}{t^2} - {{3{\pi ^2}} \over 4}\,.\,t + {{{\pi ^3}} \over 8}$

This is a quadratic equation of t.

Here, coefficient of t² term is ${{3\pi } \over 2}$ which is > 0.

$\therefore$ It is a upward parabola.

Now, $f'(t) = 3\pi t - {{3{\pi ^2}} \over 4}$

$f''(t) = 3\pi > 0$

$\therefore$ $3\pi t - {{3{\pi ^2}} \over 4} = 0$

$ \Rightarrow t = {\pi \over 4}$ (minima)

$\therefore$ vertex of graph at ${\pi \over 4}$

$\therefore$ Minimum value at ${\pi \over 4}$ and maximum value at $-$${\pi \over 2}$.

$\therefore$ $f\left( {{\pi \over 4}} \right) = {{{\pi ^3}} \over {64}} + {\left( {{\pi \over 2} - {\pi \over 4}} \right)^3} = {{{\pi ^3}} \over {32}}$

$f\left( { - {\pi \over 2}} \right) = - {{{\pi ^3}} \over 8} + {\pi ^3}$

$ = {{7{\pi ^3}} \over 8}$

$\therefore$ $k{\pi ^3} \in \left[ {{{{\pi ^3}} \over {32}},\,{{7{\pi ^3}} \over 8}} \right)$

$ \Rightarrow k \in \left[ {{1 \over {32}},\,{7 \over 8}} \right)$