Inverse Trigonometric Functions

$ \begin{aligned} & \theta=\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right) +\tan ^{-1}\left(\frac{1}{21}\right)+\tan ^{-1}\left(\frac{1}{31}\right) \\ & \quad=\sum_{k=1}^5 \tan ^{-1}\left(\frac{1}{k^2+k+1}\right) \end{aligned} $

$ \begin{aligned}and\,\,\, & \tan ^{-1}(k+1)-\tan ^{-1}(k)=\tan ^{-1}\left(\frac{1}{k^2+k+1}\right) \\ & \therefore \theta=\sum_{k=1}^5\left(\tan ^{-1}(k+1)-\tan ^{-1}(k)\right. \\ & =\tan ^{-1}(2)-\tan ^{-1}(1)+\tan ^{-1}(3) \\ & -\tan ^{-1}(2)+\tan ^{-1}(4)-\tan ^{-1}(3) \\ & +\tan ^{-1}(5)-\tan ^{-1}(4)+\tan ^{-1}(6) \\ & -\tan ^{-1}(5) \\ & =\tan ^{-1}(6)-\tan ^{-1}(1) \\ & =\tan ^{-1}\left(\frac{6-1}{1+6 \times 1}\right)=\tan ^{-1}\left(\frac{5}{7}\right) \\ & \Rightarrow \tan \theta=\frac{5}{7} \\ & \text { } \end{aligned} $

$\because \operatorname{coth}^{-1} y=\tanh ^{-1}\left(\frac{1}{y}\right)$

$ \begin{aligned} & \text { so, } \tanh ^{-1} x=\tanh ^{-1}\left(\frac{1}{y}\right) \\ & \Rightarrow \quad x=\frac{1}{y} \\ & \Rightarrow \quad x y=1 \end{aligned} $

Therefore, $\tan ^{-1}(x y)=\tan ^{-1}(1)=\frac{\pi}{4}$

$\because f(x)=2+\left|\sin ^{-1} x\right|$

Domain of $\sin ^{-1} x$ is $x \in[-1,1]$

So, the domain of $f(x)$ is $[-1,1]$

and $\left|\sin ^{-1} x\right|$ is not differentiable at $x=0$

So, the set A where $f^{\prime}(x)$ exists is

$ (-1,1)-\{0\} $

Hence, $f(x)$ is differentiable at all points in $(-1,0) \cup(0,1)$

$\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$

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

Taking $\cos$ both sides,

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

$\because x$ is positive

∴ This case is not possible

∴ Only one solution.

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \tan \left(2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}\right) \\ & =\tan \left(\tan ^{-1}\left(\frac{2 / 3}{1-1 / 9}\right)+\tan ^{-1} \frac{1}{7}\right)\left\{\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^2}\right\} \\ & =\tan \left[\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{1}{7}\right)\right] \\ & =\tan \left[\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \cdot \frac{1}{7}}\right)\right] \left\{\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y \\ =\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \end{array}\right\} \\ & =\tan \left(\tan ^{-1}\left(\frac{25}{25}\right)=1\right. \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \tanh ^{-1}\left(\frac{1}{3}\right)+\operatorname{coth}^{-1}(3) \\ & =\tanh ^{-1}\left(\frac{1}{3}\right)+\tanh ^{-1}\left(\frac{1}{3}\right) \\ & =2 \tanh ^{-1}\left(\frac{1}{3}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & =\sinh ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\frac{1}{9}}\right)=\sinh ^{-1}\left(\frac{2 / 3}{8 / 9}\right) \\ & \quad\left\{\because 2 \tanh ^{-1} x=\sinh ^{-1}\left(\frac{2 x}{1-x^2}\right)\right\} \\ & =\sinh ^{-1}\left(\frac{3}{4}\right) \end{aligned} $

$ \begin{aligned} &\text {Given, } y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\\ &\begin{aligned} & \because \quad \sin ^{-1} \frac{2 x}{1+x^2}=2 \tan ^{-1} x \\ & \therefore \quad y=2 \tan ^{-1} x \Rightarrow \frac{d y}{d x}=\frac{2}{1+x^2} \\ & \Rightarrow \frac{d^2 y}{d x^2}=\frac{(-2) 2 x}{\left(1+x^2\right)^2}=-\frac{4 x}{\left(1+x^2\right)^2} \\ & \left.\Rightarrow \frac{d^2 y}{d x^2}\right|_{x=2}=-\frac{8}{25}=k \\ & \therefore 25 k=-8=(-2)^3 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &f(x)=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right), g(x)=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \end{aligned} $

$ \begin{array}{ll} \text { put, } x=\cos \theta & x=\tan \phi \\ f(x)=\sec ^{-1}\left(\frac{1}{\cos 2 \theta}\right) & g(x)=\tan ^{-1}\left(\frac{\sec \phi-1}{\tan \phi}\right) \end{array} $

$ \begin{aligned} & =\sec ^{-1}(\sec 2 \theta) \\ & =2 \theta \\ & =2 \cos ^{-1} x \\ & \Rightarrow f^{\prime}(x)=\frac{-2}{\sqrt{1-x^2}} \end{aligned} $

$ \begin{aligned} & =\tan ^{-1}\left(\frac{1-\cos \phi}{\sin \phi}\right) \\ & =\tan ^{-1}\left(\frac{2 \sin ^2 \phi / 2}{2 \sin \phi / 2 \cos \phi / 2}\right) \\ & =\tan ^{-1}(\tan \phi / 2) \\ & =\phi / 2 \\ & =\frac{1}{2} \tan ^{-1} x \\ & \Rightarrow g^{\prime}(x)=\frac{1}{2\left(1+x^2\right)} \end{aligned} $

$ \begin{aligned} &\text { Differentiating } f(x) \text { w.r.t } g(x) \text { is }\\ &=\frac{f^{\prime}(x)}{g^{\prime}(x)}=\frac{-4\left(1+x^2\right)}{\sqrt{1-x^2}} \end{aligned} $

$ A=\left\{x \in R / \sin ^{-1}\left(\sqrt{x^2+x+1}\right) \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right\} $

and

$ B=\left\{y \in R / y=\sin ^{-1}\left(\sqrt{x^2+x+1}\right), x \in A\right\} $

Since, domain of $\sin ^{-1}(x)$ is $[-1,1]$

So, domain of $\sin ^{-1} \sqrt{x^2+x+1}$ is real, so

so

$ \begin{aligned} & \sqrt{x^2+x+1} \in[-1,1] \\ \Rightarrow \quad & \sqrt{x^2+x+1} \in[0,1] \end{aligned} $

( $\because$ square root is always non-negative)

$ \begin{array}{ll} \Rightarrow & x^2+x+1 \leq 1 \\ \Rightarrow & x^2+x \leq 0 \\ \Rightarrow & x(x+1) \leq 0 \end{array} $

So, $x \in[-1,0]$ for set $A$

Thus, $A=[-1,0]$

Now, for set $B$, since $x \in A=[-1,0]$, the range values for $\sqrt{x^2+x+1}$ for $x \in[-1,0]$ we will find.

$ \begin{aligned} \text { At } x & =-1, \sqrt{(-1)^2-1+1} \\ & =\sqrt{1-1+1}=1 \\ \text { At } x & =0, \sqrt{0^2+0+1}=\sqrt{0+1}=1 \\ \text { At } x & =-\frac{1}{2}, \sqrt{\left(\frac{-1}{2}\right)^2-\frac{1}{2}+1} \\ & =\sqrt{\frac{1}{4}-\frac{1}{2}+1} \\ & =\sqrt{\frac{3}{4}} \end{aligned} $

So, $\sqrt{x^2+x+1} \in\left[\sqrt{\frac{3}{4}}, 1\right]=\left[\frac{\sqrt{3}}{2}, 1\right]$

Thus, $y=\sin ^{-1}\left(\sqrt{x^2+x+1}\right)$

$ \begin{aligned} & \in\left[\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right), \sin ^{-1}(1)\right] \\ & =\left[\frac{\pi}{3}, \frac{\pi}{2}\right] \end{aligned} $

Hence, $B=\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$

$\Rightarrow$ Now, $A \cap B=[-1,0] \cap\left[\frac{\pi}{3}, \frac{\pi}{2}\right]=\phi$

So, $A \cap B \neq \phi$ is false.

$\Rightarrow A \cap B^C=[-1,0] \cap\left[\frac{\pi}{3}, \frac{\pi}{2}\right]^C$ and the intersection with any set outside this cannot be $[0,1]$,

So, $A \cap B^C=[0,1]$ is false.

$\Rightarrow A^C$ is everything except $[-1,0]$.

Since $B=(\pi / 3, \pi / 2) \subset R /(-1,0)$

$\Rightarrow \quad A^C \cap B$ is true.

$\Rightarrow \quad A \cup B=R-\left\{[-1,0] \cup\left[\frac{\pi}{3}, \frac{\pi}{2}\right]\right\}$

So, union of $A$ and $B$ is the complement of their own union.

That's impossible.

Thus, $A^C \cap B=\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$, option (c) is correct.

Given function,

$ f(x)=\sqrt{\log _e\left(\frac{1}{x^2-4 x+4}\right)}+\sin ^{-1}\left(x^2-2\right) $

Since, $\log$ is defined for positive values only.

So, $\frac{1}{x^2-4 x+4}>0$

Also since, $x^2-4 x+4=(x-2)^2 \geq 0$ this holds for $x \neq 2$

$ \Rightarrow \quad \ln \left(\frac{1}{(x-2)^2}\right) \geq 0 $

Now, we know that $\ln (t)$ is strictly increasing for $t>0$ and $\ln (1)=0$, the inequality

$\ln \frac{1}{(x-2)^2} \geq 0$ is equivalent to

$ \frac{1}{(x-2)^2} \geq 1 \Leftrightarrow(x-2)^2 \leq 1 $

$ \Rightarrow \quad|x-2| \leq 1 $

i.e., $\quad 1 \leq x \leq 3$

But $\log$ is undefined for $x=2$

$ \therefore x \in[1,3]-\{2\} $

For $\sin ^{-1}\left(x^2-2\right)$ to be real, $x^2-2 \in[-1,1]$

$ \begin{array}{ll} \Rightarrow & -1 \leq x^2-2 \leq 1 \\ \Rightarrow & 1 \leq x^2 \leq 3 \Rightarrow 1 \leq|x| \leq \sqrt{3} \\ \Rightarrow & x \in[-\sqrt{3},-1] \cup[1, \sqrt{3}] \Rightarrow x \in[1, \sqrt{3}] \end{array} $

Given,

$ \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\}, b>a $

Let $\tan ^{-1} \frac{a}{\sqrt{b^2-a^2}}=\theta \Rightarrow \tan \theta=\frac{a}{\sqrt{b^2-a^2}}$

$ \sec \theta=\frac{b}{\sqrt{b^2-a^2}} $

$ \begin{aligned} \therefore \cot \left(\cos ^{-1} x\right) & =\sec \left\{\tan ^{-1} \frac{a}{\sqrt{b^2-a^2}}\right\} \\ \sec \theta & =\frac{b}{\sqrt{b^2-a^2}} \\ \Rightarrow \quad \cos ^{-1} x & =\cot ^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right) \end{aligned} $

$ \begin{aligned} & \text { Again, let } \cot ^{-1}\left(\frac{b}{\sqrt{b^2-a^2}}\right)=\phi \\ & \Rightarrow \quad \cot \phi=\frac{b}{\sqrt{b^2-a^2}} \end{aligned} $

$ \Rightarrow \quad \cos \phi=\frac{b}{\sqrt{b^2-a^2}} $

Now, $\cos ^{-1} x=\phi$

$ \begin{array}{ll} \Rightarrow & x=\cos \phi=\frac{b}{\sqrt{2 b^2-a^2}} \\ \therefore & x=\frac{b}{\sqrt{2 b^2-a^2}} \end{array} $

$ \begin{aligned} & \text { Given, } \sin h^{-1} x=\log 3 \\ & \qquad \begin{aligned} & \cos h^{-1} y=\log \frac{3}{2} \\ \because & \sin h^{-1} x=\log (3) \\ \Rightarrow \quad x= & \sin h(\log 3) \\ = & \frac{e^{\log 3}-e^{-\log 3}}{2} \\ = & \frac{3-\frac{1}{3}}{2}=\frac{8}{6}=\frac{4}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { And } \cos h^{-1} y=\log \left(\frac{3}{2}\right) \\ & \Rightarrow \quad y=\cosh \left(\log \frac{3}{2}\right) \end{aligned} $

$ \begin{gathered} \Rightarrow \quad \frac{e^{\log \left(\frac{3}{2}\right)}+e^{-\log \left(\frac{3}{2}\right)}}{2}=\frac{\frac{3}{2}+\frac{2}{3}}{2}=\frac{13}{12} \\ x-y=\frac{4}{3}-\frac{13}{12} \\ =\frac{16-13}{12}=\frac{3}{12}=\frac{1}{4} \end{gathered} $

Now, we know that

$ \tan h^{-1}(z)=\frac{1}{2} \ln \left(\frac{1+z}{1-z}\right) $

Now, $z=x-y$

So, $\tan h^{-1}(x-y)=\frac{1}{2} \ln \left(\frac{1+x-y}{1-(x-y)}\right)$

$ \begin{aligned} & =\frac{1}{2} \ln \left(\frac{1+x-y}{1-x+y}\right) \\ & =\frac{1}{2} \ln \left(\frac{1+\frac{1}{4}}{1-\frac{1}{4}}\right)=\frac{1}{2} \ln \left(\frac{5 / 4}{3 / 4}\right) \\ & =\frac{1}{2} \ln \left(\frac{5}{3}\right)=\ln \left(\sqrt{\frac{5}{3}}\right) \end{aligned} $

So, $\tan h^{-1}(x-y)=\ln \left(\sqrt{\frac{5}{3}}\right)$

$ \begin{aligned} & \text { } \tan ^{-1}(1)+\frac{1}{2} \cos ^{-1}\left(x^2\right) \\ & -\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=0 \\ & \because \tan ^{-1}(1)=\pi / 4 \text { and let } \theta=\cos ^{-1}\left(x^2\right) \\ & \Rightarrow x^2=\cos \theta \\ & \text { and } 1+x^2=1+\cos x, 1-x^2=1-\cos x \\ & =\frac{\pi}{4}+\frac{\theta}{2}-\tan ^{-1}\left(\frac{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}\right) \\ & =\frac{\pi}{4}+\frac{\theta}{2}-\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta / 2+\sqrt{2} \sin \theta / 2}{\sqrt{2} \cos \theta / 2-\sqrt{2} \sin \theta / 2}\right) \\ & =\frac{\pi}{4}+\frac{\theta}{2}-\tan ^{-1}\left(\frac{\cos \theta / 2+\sin \theta / 2}{\cos \theta / 2-\sin \theta / 2}\right) \\ & \because \frac{\cos x+\sin x}{\cos x-\sin x}=\tan \left(\frac{\pi}{4}+x\right) \\ & =\frac{\pi}{4}+\frac{\theta}{2}-\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right) \\ & =\frac{\pi}{4}+\frac{\theta}{2}-\frac{\pi}{4}-\frac{\theta}{2}=0 \end{aligned} $

Hence, the identify is true for all $x$ Infinite many solutions.

$ \begin{aligned} & \text { } \because \tanh ^{-1}(x)=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)(\text { for }|x|<1) \\ & \qquad \begin{aligned} \text { so, } \tanh ^{-1}(\sin \theta) & =\frac{1}{2} \ln \left(\frac{1+\sin \theta}{1-\sin \theta}\right) \\ \frac{1+\sin \theta}{1-\sin \theta} & =\frac{(1+\sin \theta)^2}{\left(1-\sin ^2 \theta\right)}=\left(\frac{1+\sin \theta}{\cos \theta}\right)^2 \\ & =\frac{1}{2} \ln \left(\frac{1+\sin \theta}{\cos \theta}\right)^2 \\ & =\ln \left(\frac{1+\sin \theta}{\cos \theta}\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { and } \cosh ^{-1}(x)=\ln \left(x+\sqrt{x^2-1}\right) \\ & \begin{aligned} \therefore \cosh ^{-1}(\sec \theta) & =\ln \left(\frac{1}{\cos \theta}+\sqrt{\frac{1-\cos ^2 \theta}{\cos ^2 \theta}}\right) \\ & =\ln \left(\frac{1+\sin \theta}{\cos \theta}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} \end{aligned}\\ &\text { By comparing Eqs. (i) and (ii), we get }\\ &\tanh ^{-1}(\sin \theta)=\cosh ^{-1}(\sec \theta) \end{aligned} $

$ \begin{aligned} &\text { } \because \sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\\ &\begin{aligned} & \therefore f(x)=\tan ^{-1}\left(\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\right) \\ & \text { Now, } f^{\prime}(x)=\frac{1}{1+\left(\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\right)^2} \\ & \quad\left(\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)\right) \end{aligned} \end{aligned} $

$ =\frac{\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)}{1+2 \sin ^2\left(x+\frac{\pi}{4}\right)} \rightarrow \text { always }>0 $

for increasing $\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)>0$

$ \Rightarrow \quad \cos \left(x+\frac{\pi}{4}\right)>0 $

So, $\quad \frac{-\pi}{2}+2 k \pi < x+\frac{\pi}{4}<\frac{\pi}{2}+2 k \pi$

$ \Rightarrow \quad \frac{-3 \pi}{4}+2 k \pi < x <\frac{\pi}{4}+2 k \pi $

for the principal interval,

$ \text { let } k=0 $

Thus, the interval is

$ x \in\left(\frac{-3 \pi}{4}, \frac{\pi}{4}\right) $

Given, $f(x)=\cos ^{-1}\left(\frac{3}{\sqrt{9 x^2-12 x+22}}\right)$

Since, $9 x^2-12 x+22=9$

$ \begin{array}{ll} & {[\underbrace{\left(x-\frac{2}{3}\right)^2+2}_{\geq 2}] \geq 18} \\ \Rightarrow & \sqrt{9\left(x-\frac{3}{2}\right)^2+18} \geq \sqrt{18} \\ \Rightarrow & 0<\frac{1}{\sqrt{9 x^2-12 x+22}} \leq \frac{1}{\sqrt{18}} \\ \Rightarrow & 0<\frac{3}{\sqrt{9 x^2-12 x+22}} \leq \frac{3}{\sqrt{18}} \\ \Rightarrow & 0<\frac{3}{\sqrt{9 x^2-12 x+22}} \leq \frac{1}{\sqrt{2}} \end{array} $

$ \begin{gathered} \Rightarrow \quad \cos ^{-1} 0>\cos ^{-1}\left(\frac{3}{\sqrt{9 x^2-12 x+22}}\right) \\ \geq \cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ \Rightarrow \quad \frac{\pi}{4} \leq f(x)<\frac{\pi}{2} \end{gathered} $

As we know that,

$ \cot ^{-1} x=\frac{\pi}{2}-\tan ^{-1} x $

So, given equation reduces to

$ \begin{aligned} & 2\left(\frac{\pi}{2}-\tan ^{-1}\left(x^2+2 x+k\right)\right) \\ & \quad=\pi-3 \tan ^{-1}\left(x^2+2 x+k\right) \\ & \Rightarrow \quad \tan ^{-1}\left(x^2+2 x+k\right)=0 \\ & \Rightarrow \quad x^2+2 x+k=0 \end{aligned} $

Given Eq. (i) has two distinct real solution. So, its discriminant value is always greater than zero.

So, (4) $-4 k>0$

$ \Rightarrow \quad k<1 \Rightarrow k \in(-\infty, 1) $

$ \text { } \begin{aligned} & \because \operatorname{sech}^{-1}(x)=\ln \left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right) \\ & \therefore \operatorname{sech}^{-1}(\sin \alpha)=\ln \left(\frac{1}{\sin \alpha}+\sqrt{\frac{1}{\sin ^2 \alpha}}-1\right) \\ & =\ln \left(\frac{1}{\sin \alpha}+\sqrt{\frac{1-\sin ^2 \alpha}{\sin ^2 \alpha}}\right) \\ & =\ln \left(\frac{1}{\sin \alpha}+\frac{\cos \alpha}{\sin \alpha}\right) \\ & =\ln \left(\frac{1+\cos \alpha}{\sin \alpha}\right)=\ln \left(\frac{2 \cos ^2 \frac{\alpha}{2}}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}\right) \\ & =\ln \left(\cot \frac{\alpha}{2}\right) \end{aligned} $

Given $y=\log \left(\sec \left(\tan ^{-1} x\right)\right), x>0$

$ \begin{aligned} & =\log \left(\sec \left(\sec ^{-1} \frac{\sqrt{1+x^2}}{1}\right)\right) \\ & y=\log \left(\sqrt{1+x^2}\right) \end{aligned} $

$ \begin{aligned} & \therefore \text { On differentiate w.r.t. } x \text {, } \\ & \qquad \frac{d y}{d x}=\frac{1}{\sqrt{1+x^2}} \times \frac{1}{2 \sqrt{1+x^2}} \times(2 x)=\frac{x}{\left(1+x^2\right)^2} \\ & \left.\therefore \frac{d y}{d x}\right|_{x=1}=\frac{1}{1+(1)^2}=\frac{1}{2} \end{aligned} $

Given,

$ \begin{aligned} & y=\sin ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) \\ \Rightarrow & y=\sin ^{-1}\left(\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^2}{(\sqrt{1+\sin x})^2-(\sqrt{1-\sin x})^2}\right) \\ = & \sin ^{-1}\left(\frac{1+\sin x+1-\sin x+2 \sqrt{1-\sin ^2 x}}{1+\sin x-1+\sin x}\right) \\ = & \sin ^{-1}\left(\frac{2+2|\cos x|}{2 \sin x}\right)=\sin ^{-1}\left(\frac{1-\cos x}{\sin x}\right)\binom{\because \frac{-3 \pi}{2}< x < \frac{-\pi}{2}}{\therefore|\cos x|=-\cos x} \\ = & \sin ^{-1}\left(\frac{2 \sin ^2\left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}\right)=\sin ^{-1}\left(\tan \frac{x}{2}\right) \\ \therefore & \frac{d y}{d x}=\frac{1}{\sqrt{1-\tan ^2\left(\frac{x}{2}\right)}} \times \sec ^2\left(\frac{x}{2}\right) \times \frac{1}{2} \\ & =\frac{\left|\cos ^{\frac{x}{2}}\right|}{\sqrt{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}} \times \frac{1}{\left|\cos \frac{x}{2}\right|^2} \times \frac{1}{2}=\frac{\left|\sec \frac{x}{2}\right|}{2 \sqrt{\cos x}} \end{aligned} $

$ \begin{aligned} & \text { } \frac{1}{2} \sin ^{-1}\left(\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right)=\tan ^{-1} x \\ & \Rightarrow \frac{1}{2} \sin ^{-1}\left(\frac{3 \cdot \frac{2 \tan \theta}{1+\tan ^2 \theta}}{5+4\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)}\right)=\tan ^{-1} x \end{aligned} $

$ \begin{aligned} & \Rightarrow \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^2 \theta}\right)=2 \tan ^{-1} x \\ & \Rightarrow \sin ^{-1}\left(\frac{2 \cdot\left(\frac{1}{3} \tan \theta\right)}{1+\left(\frac{\tan \theta}{3}\right)^2}\right)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \\ & \Rightarrow x=\frac{1}{3} \tan \theta \end{aligned} $

$\operatorname{sech}^{-1} x=\log 2$

and $\operatorname{cosech}^{-1} y=-\log 3$

$ \begin{aligned} & \operatorname{sech}^{-1} x=\log 2 \\ & \Rightarrow \quad \log \left(\frac{1+\sqrt{1-x^2}}{x}\right)=\log 2 \\ & \Rightarrow \quad \frac{1+\sqrt{1-x^2}}{x}=2 \\ & \Rightarrow \quad 1+\sqrt{1-x^2}=2 x \\ & \quad \sqrt{1-x^2}=2 x-1 \end{aligned} $

Squaring both sides, we get

$ \begin{aligned} & \Rightarrow \quad 1-x^2=4 x^2+1-4 x \\ & \Rightarrow \quad 5 x^2-4 x=0 \\ & \Rightarrow \quad x(5 x-4)=0 \\ & \quad x=\frac{4}{5} \\ & \because \operatorname{cosech}^{-1} y=-\log 3 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad \log \left(\frac{1+\sqrt{1+y^2}}{y}\right)=\log \frac{1}{3} \\ & \Rightarrow \quad \frac{1+\sqrt{1+y^2}}{y}=\frac{1}{3} \\ & \Rightarrow \quad 1+\sqrt{1+y^2}=\frac{y}{3} \\ & \Rightarrow \quad \sqrt{1+y^2}=\frac{y}{3}-1 \end{aligned}\\ &\text { Squaring both sides, we get } \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 1+y^2=\frac{y^2}{9}+1-\frac{2}{3} y \\ & \Rightarrow \quad \frac{8 y^2}{9}+\frac{2}{3} y=0 \\ & \Rightarrow \quad \frac{2}{3} y\left(\frac{4 y}{3}+1\right)=0 \\ & \Rightarrow \quad y=\frac{-3}{4} \\ & \Rightarrow \quad x+y=\frac{4}{5}-\frac{3}{4}=\frac{16-15}{20} \\ & \Rightarrow \quad x+y=\frac{1}{20} \end{aligned} $

$ \begin{aligned} & y=\tan ^{-1}\left(\frac{x}{1+2 x^2}\right)+\tan ^{-1}\left(\frac{x}{1+6 x^2}\right) \\ & \Rightarrow y=\tan ^{-1} 2 x-\tan ^{-1} x+\tan ^{-1} 3 x \\ & \Rightarrow y=\tan ^{-1} 3 x-\tan ^{-1} x \\ & \Rightarrow \frac{d y}{d x}=\frac{1 \times 3}{1+9 x^2}-\frac{1}{1+x^2} \\ & \Rightarrow \frac{d y}{d x}=\frac{3}{9 x^2+1}-\frac{1}{x^2+1} \end{aligned} $

$ \begin{aligned} & \text { } f(x)=\cos ^{-1}(-x)+\sin ^{-1}(-x)\left.+\operatorname{cosec}^{-1}(x)\right] \\ & \text { Domain }=(-1,1) \\ & \Rightarrow \quad f(x)=\pi-\cos ^{-1} x-\sin ^{-1} x \quad+\operatorname{cosec}^{-1} x \\ & \Rightarrow \quad f(x)=\pi-\frac{\pi}{2}+\operatorname{cosec}^{-1} x \\ & \Rightarrow \quad f(x)=\frac{\pi}{2}+\operatorname{cosec}^{-1} x \end{aligned} $

Domain of $\operatorname{cosec}^{-1} x$

$ (-\infty,-1] \cup[1, \infty) $

The range of $\operatorname{cosec}^{-1} x$ is

$ \begin{array}{r} {\left[-\frac{\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right]} \\ \because f(x)=\frac{\pi}{2}+\operatorname{cosec}^{-1} x \end{array} $

$\therefore$ Range of

$ \begin{aligned} f(x) & =\left[\frac{\pi}{2}-\frac{\pi}{2}, \frac{\pi}{2}+0\right) \cup\left(0+\frac{\pi}{2}, \frac{\pi}{2}+\frac{\pi}{2}\right] \\ & =\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right] \end{aligned} $

$ \text { In } \triangle A D B \text {, } $

$ \begin{aligned} &\begin{aligned} \Rightarrow \quad \tan 30^{\circ}=\frac{x}{10 \sqrt{3}} & \\ x & =10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ \triangle C D B, \tan 60^{\circ} & =\frac{x+y}{10 \sqrt{3}} \\ x+y & =30 \end{aligned}\\ &\therefore \text { Sum of height }=10+30=40 \end{aligned} $

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

So, $x(x+1)=0$ or $x^2+x+2=0$ (rejected)

$ \Rightarrow x(x+1)=0 \Rightarrow x=0,-1 $

$\because$ Only two real solution.

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y=\tan h^{-1} \sqrt{\frac{1-x}{1+x}} \\ & \because \quad \frac{d}{d x} \tan h^{-1} x=\frac{1}{1-x^2} \\ & \therefore \quad \frac{d y}{d x}=\frac{1}{1-\frac{1-x}{1+x}} \times \frac{1}{2} \sqrt{\frac{1+x}{1-x}} \times \frac{(1+x)(-1)-(1-x)}{(1+x)^2} \\ & =\frac{1+x}{1+x-1+x} \times \frac{1}{2} \frac{\sqrt{1+x}}{\sqrt{1-x}} \times \frac{-2}{(1+x)^2} \\ & =-\frac{1+x}{2 x} \times \frac{1}{(1+x) \sqrt{(1+x)(1-x)}} \\ & =-\frac{1}{2 x \sqrt{1-x^2}} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \tan ^{-1} \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}+\tan ^{-1} \frac{1}{\sqrt{5}} \\ & \begin{aligned} = & \tan ^{-1}\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{15}+1}\right)+\tan ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\ = & \tan ^{-1}\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{15}+1} \times \frac{\sqrt{15}-1}{\sqrt{15}-1}\right)+\tan ^{-1} \frac{1}{\sqrt{5}} \\ = & \tan ^{-1}\left(\frac{5 \sqrt{3}-3 \sqrt{5}-\sqrt{5}+\sqrt{3}}{14}\right) \quad+\tan ^{-1} \frac{1}{\sqrt{5}} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\tan ^{-1}\left(\frac{6 \sqrt{3}-4 \sqrt{5}}{14}\right)+\tan ^{-1} \frac{1}{\sqrt{5}} \\ & =\tan ^{-1}\left(\frac{3 \sqrt{3}-2 \sqrt{5}}{7}\right)+\tan ^{-1} \frac{1}{\sqrt{5}} \\ & =\tan ^{-1}\left(\frac{\frac{3 \sqrt{3}-2 \sqrt{5}}{7}+\frac{1}{\sqrt{5}}}{1-\frac{3 \sqrt{3}-2 \sqrt{5}}{7} \times \frac{1}{\sqrt{5}}}\right) \\ & =\tan ^{-1}\left(\frac{3 \sqrt{15}-10+7}{7 \sqrt{5}-3 \sqrt{3}+2 \sqrt{5}}\right) \\ & =\tan ^{-1}\left(\frac{3 \sqrt{15}-3}{9 \sqrt{5}-3 \sqrt{3}}\right) \\ & =\tan ^{-1}\left(\frac{\sqrt{15}-1}{3 \sqrt{5}-\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\right) \\ & =\tan ^{-1}\left(\frac{3 \sqrt{5}-\sqrt{3}}{3 \sqrt{5}-\sqrt{3}} \times \frac{1}{\sqrt{3}}\right) \\ & =\tan ^{-1} \frac{1}{\sqrt{3}}=\frac{\pi}{6} \end{aligned} $

$ \begin{aligned} & \sec ^{-1} \frac{1}{2 x^2-1}=\cos ^{-1}\left(2 x^2-1\right)=2 \cos ^{-1} x \\ & \because \quad f(x)= \\ & \Rightarrow \quad f^{\prime}(x)=-\frac{2 \cos ^{-1} x}{\sqrt{1-x^2}} \\ & \Rightarrow \quad g(x)=\sqrt{1-x^2} \\ & \Rightarrow \quad g^{\prime}(x)=\frac{1}{2 \sqrt{1-x^2}} \times(-2 x) \\ & =-\frac{x}{\sqrt{1-x^2}} \\ & \Rightarrow \quad \frac{d f}{d g}=\frac{-\frac{2}{\sqrt{1-x^2}}}{-\frac{x}{\sqrt{1-x^2}}}=\frac{2}{x} \\ & \Rightarrow\left(\frac{d f}{d g}\right)_{\text {at } x=\frac{1}{2}}=\frac{2}{\frac{1}{2}}=4 \end{aligned} $

We have, $0

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

Let $x=\sin \theta$

$ \begin{aligned} \alpha & =\sin ^{-1}(\sin \theta)+\cos ^{-1} \\ & =\theta+\cos ^{-1}\left(\cos \left(\frac{\pi}{6}-\theta\right)\right) \\ & =\theta+\frac{\pi}{6}-\theta=\frac{\pi}{6} \\ \tan \alpha & +\cot \alpha \Rightarrow \tan \frac{\pi}{6}+\cot \frac{\pi}{6} \\ & =\frac{1}{\sqrt{3}}+\sqrt{3}=\frac{4}{\sqrt{3}} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \cot \left[\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right) \\ & =\sum_{n=1}^{50} \tan ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\sum_{n=1}^{50}\left[\tan ^{-1}(n+1)-\tan ^{-1} n\right] \\ & =\tan ^{-1} 51-\tan ^{-1} 1 \\ & =\tan ^{-1}\left(\frac{51-1}{1+51 \cdot 1}\right)=\tan ^{-1}\left(\frac{50}{52}\right) \end{aligned}\\ &\text { Now, from Eq. (i), we get }\\ &\begin{aligned} \Rightarrow \quad \cot \left(\tan ^{-1}\left(\frac{50}{52}\right)\right) & =\cot \left(\cot ^{-1}\left(\frac{52}{50}\right)\right) \\ & =\frac{52}{50}=\frac{26}{25} \end{aligned} \end{aligned} $

To find the value of $ x $, we need to equate:

$ \sin \left(2 \tan^{-1} \frac{3}{4}\right) = \cos \left(2 \tan^{-1} x\right). $

Step-by-step Solution:

1. Left-hand side (LHS):

Let $\theta = \tan^{-1} \frac{3}{4}$, which implies $\tan \theta = \frac{3}{4}$.

We have:

$ \sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2 \times \frac{3}{4}}{1 + \left(\frac{3}{4}\right)^2} $

Calculating further:

$ = \frac{\frac{3}{2}}{1 + \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{25}{16}} = \frac{24}{25}. $

2. Right-hand side (RHS):

Let $ z = \tan^{-1} x $, hence $ x = \tan z $.

Therefore:

$ \cos(2z) = \frac{1 - \tan^2 z}{1 + \tan^2 z} = \frac{1 - x^2}{1 + x^2}. $

3. Equating LHS to RHS:

$ \frac{24}{25} = \frac{1 - x^2}{1 + x^2}. $

Cross-multiplying gives:

$ 24 + 24x^2 = 25 - 25x^2. $

Simplifying:

$ 49x^2 = 1, $

which leads to:

$ x^2 = \frac{1}{49} = \frac{1}{7^2}, $

thus:

$ x = \frac{1}{7}. $

Given that

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

$ \text { Let } t=\frac{1+x^2}{2 x} \text { and } u=\frac{2 x}{1+x^2} $

$ \because \quad t=\frac{1}{u} $

$ \therefore \quad t \cdot u \equiv 1 $

We know that

$ \sin ^{-1}(a)+\cos ^{-1}(a)=\frac{\pi}{2}, a \in[-1,1] $

Thus, $\sin ^{-1}(t)+\cos ^{-1}(u)=\frac{\pi}{2}$

So, the range of the function

$ f(x)=\sin ^{-1}\left(\frac{1+x^2}{2 x}\right)+\cos ^{-1}\left(\frac{2 x}{1+x^2}\right) \text { is }\left\{\frac{\pi}{2}\right\} $

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

We know that,

$ \begin{aligned} & \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \text { where } \\ & x y<1 \end{aligned} $

Now, $\tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4}$

$\Rightarrow \quad \tan ^{-1}\left(\frac{x+2 x}{1-x \cdot 2 x}\right)=\frac{\pi}{4}$

$\Rightarrow \quad \tan ^{-1}\left(\frac{3 x}{1-2 x^2}\right)=\frac{\pi}{4}$

$\Rightarrow \quad \frac{3 x}{1-2 x^2}=\tan \frac{\pi}{4}$

$\Rightarrow \quad \frac{3 x}{1-2 x^2}=1$

$\Rightarrow \quad \frac{3 x}{1-2 x^2}=1 \quad\left[\because \tan \frac{\pi}{4}=1\right]$

$ \Rightarrow \quad 3 x=1-2 x^2 $

$\Rightarrow \quad 2 x^2+3 x-1=0$

Using quadratic formula

$ \begin{aligned} &x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\\ &\text { Where, } a=2, b=3, c=-1\\ &\begin{aligned} x & =\frac{-3 \pm \sqrt{9-4 \times 2 \times(-1)}}{4} \\ & =\frac{-3 \pm \sqrt{9+8}}{4} \\ \Rightarrow \quad x & =\frac{-3 \pm \sqrt{17}}{4} \end{aligned} \end{aligned} $

$ \text { Thus, } \quad x=\frac{-3+\sqrt{17}}{4}, \text { or } \frac{-3-\sqrt{17}}{4} $

$ \begin{aligned} &\text { }\\ &For,\,\,\,\,\,\,x=\frac{-3-\sqrt{17}}{4} \end{aligned} $

Since, $x \cdot 2 x<1 \Rightarrow 2 x^2<1$

$\Rightarrow 2\left(\frac{-3-\sqrt{17}}{4}\right)^2<1$, which is not true.

$\therefore \quad x=\frac{-3-\sqrt{17}}{4}$ is not possible

For $x=\left(\frac{-3+\sqrt{17}}{4}\right)$

Now, $\quad 2 x^2<1$

$ 2\left(\frac{-3+\sqrt{17}}{4}\right)^2<1 $

Hence, $x=\frac{\sqrt{17}-3}{4}$ is the only solution of the given equation.

We have, $2 \operatorname{coth}^{-1}(4)+\operatorname{sech}^{-1}\left(\frac{3}{5}\right)$

We know that $\operatorname{coth}^{-1}(x)=\frac{1}{2} \log \left(\frac{x+1}{x-1}\right)$

$ \begin{aligned} & \operatorname{sech}^{-1}(x)=\log \left(\frac{1+\sqrt{1-x^2}}{x}\right), 0 < x \leq 1 \\ & \therefore 2 \operatorname{coth}^{-1}(4)+\operatorname{sech}^{-1}\left(\frac{3}{5}\right) \\ & =2 \times \frac{1}{2} \log \left(\frac{4+1}{4-1}\right)+\log \left(\frac{1+\sqrt{1-\left(\frac{3}{5}\right)^2}}{\frac{3}{5}}\right) \\ & =\log \left(\frac{5}{3}\right)+\log \left(\frac{1+\frac{1}{5} \sqrt{25-9}}{\frac{3}{5}}\right) \\ & =[\log 5-\log 3]+\log \left(\frac{5+\sqrt{16}}{3}\right) \\ & =[\log 5-\log 3]+\log \left(\frac{5+4}{3}\right) \\ & =[\log 5-\log 3)+\log \left(\frac{9}{3}\right) \\ & =[\log 5-\log 3]+\log 3 \\ & =\log 5 \end{aligned} $

If $y=\sin ^{-1} x$, then

$ \left(1-x^2\right) y_2-x y_1=\text { ? } $

We have, $\Rightarrow y=\sin ^{-1} x$

On differentiating $y$ w.r.t. $x$, we get

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=y_1=\frac{d}{d x}\left(\sin ^{-1}(x)\right) \\ & \Rightarrow \quad \frac{d y}{d x}=y_1=\frac{1}{\sqrt{1-x^2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Again differentiate Eq. (i) w.r.t. $x$, we get

$ \begin{aligned} & \Rightarrow y_2=\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{1}{\sqrt{1-x^2}}\right) \\ & \Rightarrow y_2=\frac{x}{\sqrt{1-x^2}\left(1-x^2\right)} \\ & \Rightarrow y_2\left(1-x^2\right)=\frac{x}{\sqrt{1-x^2}} \end{aligned} $

From Eq. (i), we get

$ \begin{aligned} & y_2\left(1-x^2\right)=x y_1 \\ \Rightarrow \quad & y_2\left(1-x^2\right)-x y_1=0 \end{aligned} $

Given, $\cos ^{-1} 2 x+\cos ^{-1} 3 x=\frac{\pi}{3}$

$ \begin{aligned} & \cos ^{-1}\left[(2 x)(3 x)-\sqrt{1-(2 x)^2} \sqrt{1-(3 x)^2}\right]=\frac{\pi}{3} \\ & \Rightarrow 6 x^2-\sqrt{1-4 x^2} \sqrt{1-9 x^2}=\cos \frac{\pi}{3} \\ & \Rightarrow 6 x^2-\frac{1}{2}=\sqrt{\left(1-4 x^2\right)\left(1-9 x^2\right)} \end{aligned} $

On squaring both sides, we get

$ \begin{aligned} & 36 x^4+\frac{1}{4}-2 \cdot\left(6 x^2\right) \frac{1}{2}=1-9 x^2-4 x^2+36 x^4 \\ & \Rightarrow \quad \frac{1}{4}-6 x^2=1-13 x^2 \Rightarrow 7 x^2=\frac{3}{4} \\ & \Rightarrow \quad x^2=\frac{3}{28} \Rightarrow x=\frac{\sqrt{3}}{2 \sqrt{7}} \\ & 4 x^2=4 \times \frac{3}{28}=\frac{3}{7}=\frac{a}{b} \quad \text { [given] } \end{aligned} $

Thus, $a+b=3+7=10$

Given, $\theta=\sec ^{-1}(\cos h u)$

$ \begin{array}{ll} \Rightarrow & \sec \theta=\cos h u \\ \Rightarrow & u=\cos h^{-1}(\sec \theta) \end{array} $

Since, $\cos h^{-1}(x)=\log \left(x+\sqrt{x^2-1}\right)$

$ \begin{aligned} \cos h^{-1} & (\sec \theta)=\log \left(\sec \theta+\sqrt{\sec ^2 \theta-1}\right) \\ & =\log (\sec \theta+\tan \theta) \\ & =\log \left(\frac{1+\sin \theta}{\cos \theta}\right) \\ & =\log \left(\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}\right) \\ & =\log \left(\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}\right) \\ & =\log \left(\frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}\right) \end{aligned} $

$ \begin{aligned} & =\log \left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right) \\ \therefore \quad u & =\log \left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right) . \end{aligned} $

We are given:

$ \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8} = k $

This can be rewritten as:

$ \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \left[\cos(\pi - \frac{3\pi}{8})\right]^4 + \left[\cos(\pi - \frac{\pi}{8})\right]^4 = k $

Since $\cos(\pi - \theta) = -\cos(\theta)$, the expression simplifies to:

$ \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{\pi}{8} = k $

Thus, we have:

$ 2 \cos^4 \frac{\pi}{8} + 2 \cos^4 \frac{3\pi}{8} = k $

Next, using the values:

$ \cos \frac{\pi}{8} = \sqrt{\frac{2+\sqrt{2}}{4}} \quad \text{and} \quad \cos \frac{3\pi}{8} = \sqrt{\frac{2-\sqrt{2}}{4}} $

Thus:

$ 2 \left[\left(\frac{2+\sqrt{2}}{4}\right)^2 + \left(\frac{2-\sqrt{2}}{4}\right)^2\right] = k $

Expanding these squares:

$ \left(\frac{2+\sqrt{2}}{4}\right)^2 = \frac{4 + 2\sqrt{2} + 1}{16} = \frac{6 + 4\sqrt{2}}{16} $

$ \left(\frac{2-\sqrt{2}}{4}\right)^2 = \frac{4 - 2\sqrt{2} + 1}{16} = \frac{6 - 4\sqrt{2}}{16} $

Adding these together:

$ \frac{6 + 4\sqrt{2}}{16} + \frac{6 - 4\sqrt{2}}{16} = \frac{12}{16} = \frac{3}{4} $

Thus:

$ k = 2 \times \frac{3}{4} = \frac{3}{2} $

Now, we compute:

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

$ = \sin^{-1}\left(\sqrt{\frac{3}{4}}\right) + \cos^{-1}\left(\frac{1}{2}\right) $

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

$ = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} $

We are given the expression:

$ 4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{70} + \tan^{-1} \frac{1}{99} $

Let's simplify this expression step-by-step:

First, break the expression:

$ 4 \tan^{-1} \frac{1}{5} = 2 \left(\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{5}\right) $

Use the addition formula for arctan:

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

Applying this to $\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{5}$:

$ = \tan^{-1}\left(\frac{\frac{1}{5} + \frac{1}{5}}{1 - \frac{1}{5} \cdot \frac{1}{5}}\right) = \tan^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right) = \tan^{-1}\left(\frac{5}{12}\right) $

$ \text{Therefore, } 2 \left(\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{5}\right) = 2 \tan^{-1}\left(\frac{5}{12}\right) $

Now incorporate the other terms:

$ 2 \tan^{-1}\left(\frac{5}{12}\right) + \tan^{-1}\left(\frac{1}{99}\right) - \tan^{-1}\left(\frac{1}{70}\right) $

Simplify further using the arctan subtraction formula:

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

$ \tan^{-1}\left(\frac{1}{99}\right) - \tan^{-1}\left(\frac{1}{70}\right) = \tan^{-1}\left(\frac{\frac{1}{99} - \frac{1}{70}}{1 + \frac{1}{99} \cdot \frac{1}{70}}\right) $

$ = \tan^{-1}\left(\frac{-29}{6931}\right) $

Final simplification:

$ 2 \tan^{-1}\left(\frac{5}{12}\right) - \tan^{-1}\left(\frac{1}{239}\right) $

$ \tan^{-1}\left(\frac{\frac{5}{6}}{1 - \frac{25}{144}}\right) - \tan^{-1}\left(\frac{29}{6931}\right) $

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

$ \tan^{-1}\left(\frac{120}{119}\right) - \tan^{-1}\left(\frac{1}{239}\right) $

Arrive at the final form:

Using the formula:

$ \tan^{-1}\left(\frac{120 \times 239 - 119}{119 \times 239 + 120}\right) = \tan^{-1}(1) = \frac{\pi}{4} $

Thus, the simplified result of the given expression is $\frac{\pi}{4}$.

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} $

$\begin{aligned} & \tan ^{-1}(-2)-\tan ^{-1}(3)=\tan ^{-1}\left[\frac{(-2)-3}{1+(-2)(3)}\right] \\ & =\tan ^{-1}\left[\frac{-5}{-5}\right]=\tan ^{-1} 1\end{aligned}$

From options, $\tan \frac{3 \pi}{4}=-1$

$\begin{aligned} & \Rightarrow \quad \tan \left(\frac{-3 \pi}{4}\right)=1 \Rightarrow \tan ^{-1}(1)=\frac{-3 \pi}{4} \\ & \therefore \tan ^{-1}(-2)-\tan ^{-1}(3)=\frac{-3 \pi}{4} \end{aligned}$

$\begin{aligned} & x=\sin \left(2 \tan ^{-1} 2\right)=\frac{2 \cdot 3}{1+3^2}=\frac{3}{5} \\ & y=\cos \left(2 \tan ^{-1} 3\right)=\frac{\left(1-3^2\right)}{\left(1+3^2\right)}=\frac{-4}{5} \\ & z=\sec \left(3 \tan ^{-1} 4\right)=\tan ^{-1} 4=\theta \\ & \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta \\ & =4 \cdot\left(\frac{1}{17 \sqrt{17}}\right)-3\left(\frac{1}{\sqrt{17}}\right) \\ & \end{aligned}$

$=\frac{-47}{17\sqrt{17}}$

$\sec3\theta=\frac{-17\sqrt{17}}{47}$

$\therefore \quad x > y > z$.

Let $x=\cos 2 \theta$

$\begin{aligned} & \Rightarrow \frac{d \theta}{d x}=-\frac{1}{\sin 2 \theta} \\ & \therefore \frac{d}{d x}\left\{\sin ^{-1}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)\right\}=\frac{d}{d x} \sin ^2 \theta=\sin 2 \theta \frac{d \theta}{d x}=-1 \end{aligned}$

$\begin{aligned} y^{\prime} & =\left[\frac{1}{1+\left(\frac{a x-b}{b x+a}\right)^2}\right] \frac{(b x+a) a-(a x-b) b}{(b x+a)^2} \\ & =\left[\frac{1}{\left(a^2+b^2\right)\left(x^2+1\right)}\right]\left(a^2+b^2\right)=\frac{1}{x^2+1}\end{aligned}$

$\begin{aligned} & \text { } \sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0 \\ & \Rightarrow \quad 2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=\sin ^{-1}(0)=0 \\ & \Rightarrow \quad \cos ^{-1}\left(\cot \left(2 \tan ^{-1} x\right)\right)=0 \\ & \Rightarrow \quad \cot \left[\left(2 \tan ^{-1} x\right)\right]=\cos 0=1 \\ & \Rightarrow \quad \cot \left[\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]=1 \\ & \Rightarrow \quad \cot \left[\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)\right]=1 \\ & \Rightarrow \quad \frac{1-x^2}{2 x}=1 \\ & \Rightarrow \quad 1-x^2=2 x \\ & \Rightarrow \quad x^2+2 x-1=0 \\ & x=\frac{-2 \pm \sqrt{4+4}}{2}=\frac{-2 \pm 2 \sqrt{2}}{2} \\ & x=-1 \pm \sqrt{2} \\ \end{aligned}$

Given equation holds for 2 values of $x$.

$\begin{aligned} & \text { } \tan ^{-1}\left(\frac{1}{1+1 \cdot 2}\right)+\tan ^{-1}\left(\frac{1}{1+2 \cdot 3}\right)+\ldots \\ & \quad+\tan ^{-1}\left(\frac{1}{1+n(n+1)}\right)=\tan ^{-1}(x) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{2-1}{1+1 \cdot 2}\right)+\tan ^{-1}\left(\frac{3-2}{1+2 \cdot 3}\right)+\ldots \\ & \quad+\tan ^{-1}\left(\frac{n+1-n}{1+n(n+1)}\right)=\tan ^{-1}(x) \\ & \Rightarrow \quad \tan ^{-1}(2)-\tan ^{-1}(1)+\tan ^{-1}(3)-\tan ^{-1}(2)+\ldots \\ & \Rightarrow \quad \tan ^{-1}(n+1)-\tan ^{-1}(n)=\tan ^{-1}(x) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{n+1-1}{1+(n+1) \cdot 1}\right)=\tan ^{-1}(x)=\tan ^{-1}(x) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{n}{n+2}\right)=\tan ^{-1}(x) \Rightarrow x=\frac{n}{n+2}\end{aligned}$

$y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$

Substituting $x^2=\cos 2 \theta$

$\begin{aligned} \Rightarrow \quad \theta & =\frac{1}{2} \cos ^{-1}\left(x^2\right) \\ \therefore \quad y & =\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right) \\ & =\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right) \\ & =\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \\ & =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right) \Rightarrow y=\frac{\pi}{4}+\theta \\ \Rightarrow \quad y & =\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(x^2\right) \\ \therefore \quad \frac{d y}{d x} & =\frac{1}{2} \cdot\left(\frac{-2 x}{\sqrt{1-x^4}}\right)=\frac{-x}{\sqrt{1-x^4}}\end{aligned}$

$\frac{1}{k} \sec ^{-1}\left(x^k\right)=\int \frac{d x}{x \sqrt{x^4-1}}=\frac{1}{2} \int \frac{2 x d x}{x^2 \sqrt{x^4-1}}$

Let

$\begin{aligned} & \Rightarrow \quad 2 x d x=d t \\ & =\frac{1}{2} \int \frac{d t}{t \sqrt{t^2-1}}=\frac{1}{2} \sec ^{-1} t \\ & \Rightarrow \frac{1}{k} \sec ^{-1}\left(x^k\right)=\frac{1}{2} \sec ^{-1}\left(x^2\right) \Rightarrow k=2 \\ & \end{aligned}$