Inverse Trigonometric Functions

To solve the expression $\cosh \left(\sinh^{-1}(\sqrt{8}) + \cosh^{-1} 5\right)$, follow the steps outlined below:

Express the Inverse Hyperbolic Functions as Logarithms:

$ \sinh^{-1}(\sqrt{8}) = \log(\sqrt{8} + \sqrt{8 + 1}) $

Simplifying:

$ = \log(\sqrt{8} + 3) = \log(3 + \sqrt{8}) $

Similarly, for $\cosh^{-1}(5)$:

$ \cosh^{-1}(5) = \log(5 + \sqrt{5^2 - 1}) $

Simplifying:

$ = \log(5 + \sqrt{24}) = \log(5 + 2\sqrt{6}) $

Combine the Logarithms:

$ \sinh^{-1}(\sqrt{8}) + \cosh^{-1}(5) = \log((3+\sqrt{8}) \cdot (5+2\sqrt{6})) $

Evaluate the Product Inside the Logarithm:

Calculate:

$ (3+\sqrt{8})(5+2\sqrt{6}) = 3 \cdot 5 + 3 \cdot 2\sqrt{6} + \sqrt{8} \cdot 5 + \sqrt{8} \cdot 2\sqrt{6} $

$ = 15 + 6\sqrt{6} + 5\sqrt{8} + 2\sqrt{48} $

Simplifying $\sqrt{8} = 2\sqrt{2}$ and $\sqrt{48} = 4\sqrt{3}$:

$ = 15 + 6\sqrt{6} + 10\sqrt{2} + 8\sqrt{3} $

Calculate the $\cosh(x)$ Value:

Since $e^x = 15 + 10\sqrt{2} + 6\sqrt{6} + 8\sqrt{3}$, $\cosh(x)$ is given by:

$ \cosh(x) = \frac{e^x + e^{-x}}{2} $

Simplify $\cosh(x)$:

$ = \frac{1}{2}\left[(3+2\sqrt{2})(5+2\sqrt{6}) + (3-2\sqrt{2})(5-2\sqrt{6})\right] $

$ = \frac{1}{2}[30 + 16\sqrt{3}] = 15 + 8\sqrt{3} $

Thus, the expression evaluates to $15 + 8\sqrt{3}$.

To solve the expression $\tan^{-1} 2 + \tan^{-1} 3$, we can use the formula for the sum of inverse tangents:

$ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) $

Applying this formula to our problem, we have:

$ \tan^{-1} 2 + \tan^{-1} 3 = \tan^{-1} \left( \frac{2 + 3}{1 - 2 \times 3} \right) $

Simplify the fraction:

$ = \tan^{-1} \left( \frac{5}{1 - 6} \right) = \tan^{-1}(-1) $

We know that $\tan^{-1}(-1) = -\frac{\pi}{4}$. However, we typically express angles in the range $0$ to $\pi$. Thus:

$ \tan^{-1} 2 + \tan^{-1} 3 = \frac{3\pi}{4} $

Given the function $ y = \tan^{-1}\left(\frac{\sqrt{x}-x}{1+x^{\frac{3}{2}}}\right) $.

We can rewrite this expression using an identity for inverse tangent:

$ y = \tan^{-1}\left(\frac{\sqrt{x}-x}{1+\sqrt{x} \cdot x}\right) $

This can be expressed as:

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

This transformation uses the identity:

$ \tan^{-1}\left(\frac{a-b}{1+ab}\right) = \tan^{-1}(a) - \tan^{-1}(b) $

To find the derivative $ y' $ with respect to $ x $, we use the differentiation of inverse tangent functions:

$ y' = \frac{d}{dx}[\tan^{-1}(\sqrt{x}) - \tan^{-1}(x)] $

The derivative is calculated as follows:

$ y' = \frac{1}{1+(\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} - \frac{1}{1+x^2} $

Now, evaluate $ y'(1) $:

$ y'(1) = \frac{1}{1+1} \cdot \frac{1}{2\cdot1} - \frac{1}{1+1} = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} $

Thus, the value of $ y'(1) $ is $ -\frac{1}{4} $.

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}$

Given, $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1} \sqrt{\cos \alpha}=x$ ... (i)

We know that $\cot ^{-1}(\sqrt{\cos } \alpha)+\tan ^{-1} \sqrt{\cos \alpha}=\frac{\pi}{2}$ .... (ii)

On adding Eqs. (i) and (ii), we get

$\begin{aligned} & 2 \cot ^{-1} \sqrt{\cos \alpha}=\frac{\pi}{2}+x \\ & \Rightarrow \sqrt{\cos } \alpha=\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)=\frac{\cot \frac{x}{2}-1}{1+\cot \frac{x}{2}} \end{aligned}$

$\Rightarrow \sqrt{\cos \alpha}=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}$

On squaring both sides, we get

$\cos \alpha=\frac{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}$

$\begin{aligned} & \Rightarrow \cos \alpha=\frac{1-\sin x}{1+\sin x} \\ & \Rightarrow \frac{1-\tan ^2 \frac{\alpha}{2}}{1+\tan ^2 \frac{\alpha}{2}}=\frac{1-\sin x}{1+\sin x} \end{aligned}$

On applying componendo and dividendo rule, we get

$\begin{aligned} & \frac{1-\tan ^2 \frac{\alpha}{2}-1-\tan ^2 \frac{\alpha}{2}}{1-\tan ^2 \frac{\alpha}{2}+1+\tan ^2 \frac{\alpha}{2}}=\frac{1-\sin x-1-\sin x}{1-\sin x+1+\sin x} \\ \Rightarrow \quad & \tan ^2 \frac{\alpha}{2}=\sin x \\ \Rightarrow \quad & \sin x=\tan ^2 \frac{\alpha}{2} \end{aligned}$

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

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)$

$y = {\cos^{ - 1}}\left( {{7 \over 2}(1 + \cos 2x) + \sqrt {({{\sin }^2}x - 48{{\cos }^2}x)} \sin x} \right)$

$ = {\cos ^{ - 1}}((7\cos x)(\cos x) + \sqrt {1 - 49{{\cos }^2}x} \sqrt {1 - {{\cos }^2}x} )$

$ = {\cos ^{ - 1}}(\cos x) - {\cos ^{ - 1}}(7\cos x)$ [$\because$ $\cos x < 7\cos x$]

$ = x - {\cos ^{ - 1}}(7\cos x)$