Inverse Trigonometric Functions

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} $

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

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

$\begin{aligned} & \tan \left(\tan ^{-1}\left(\frac{3}{4}\right)-2 \tan ^{-1}\left(\frac{1}{2}\right)\right) \\ & \tan \left(\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{2 \times \frac{1}{2}}{1-\frac{1}{4}}\right)\right) \\ & \tan \left(\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{4}{3}\right)\right) \\ & \tan \left(\tan ^{-1}\left(\frac{3}{4}-\frac{4}{3}\right)\right) \\ & \frac{9-16}{24}=\frac{-7}{24} \end{aligned}$

$ \sin ^{-1} x-\cos ^{-1} 2 x=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} $

$ \Rightarrow \frac{\pi}{2}-\cos ^{-1} x-\cos ^{-1} 2 x=\frac{\pi}{6} $

$ \Rightarrow \cos ^{-1} x+\cos ^{-1} 2 x=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3} $

$ \Rightarrow \cos ^{-1}\left(x \cdot 2 x-\sqrt{1-x^{2}} \sqrt{1-4 x^{2}}\right)=\frac{\pi}{3} $

$ \Rightarrow 2 x^{2}-\sqrt{1-x^{2}} \sqrt{1-4 x^{2}}=\cos \frac{\pi}{3}=\frac{1}{2} $

$ \Rightarrow \sqrt{1-x^{2}} \sqrt{1-4 x^{2}}=\left(2 x^{2}-\frac{1}{2}\right)^{2} $

On solving, we get

$ x=\frac{1}{2} $

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

$ \begin{array}{l} =\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right) \\\\ =\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)=\tan ^{-1} 1=\frac{\pi}{4} \end{array} $

$ \sec h^{-1} \frac{3}{5}-\tan ^{-1} \frac{3}{5} $

Let $ \operatorname{sech}^{-1} \frac{3}{5}=x $ and $ \tan h^{-1} \frac{3}{5}=y $

$ \begin{array}{l} \sec h x=\frac{3}{5} \text { and } \tan h y=\frac{3}{5} \\\\ x=\ln 3 \quad e^{2 y}=4 \\\\ e y=2 \\\\ y=\ln 2 \end{array} $

Now, $ x-y=\ln 3-\ln 2=\ln \frac{3}{2} $

To find the domain of the function $ f(x) = \sin^{-1}\left(\log_{2}\left(\frac{x^{2}}{2}\right)\right) $, we need to determine where the expression $\log_{2}\left(\frac{x^{2}}{2}\right)$ falls within the domain of the inverse sine function, which is $[-1, 1]$.

First, ensure that:

$ -1 \leq \log_{2}\left(\frac{x^{2}}{2}\right) \leq 1 $

This inequality translates to:

$ 2^{-1} \leq \frac{x^{2}}{2} \leq 2 $

Solving the inequality:

$ \frac{1}{2} \leq \frac{x^{2}}{2} \leq 2 $

Multiply through by 2 to clear the fraction:

$ 1 \leq x^{2} \leq 4 $

This implies that $ x $ must satisfy:

$ x \in [-2, -1] \cup [1, 2] $

We also need to ensure that $\frac{x^{2}}{2} \geq 0$, which is always true since $x^2 \geq 0$.

Therefore, considering the intersection of the feasible solutions:

$ x \in [-2, -1] \cup [1, 2] $

Thus, the domain of the function is:

$ x \in [-2, -1] \cup [1, 2] $

To solve the trigonometric equation $\sin^{-1} x = 2 \sin^{-1} a$, we must consider the following:

Given that $\sin^{-1} x = 2 \sin^{-1} a$, it implies:

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

This corresponds to:

$ 2 \sin^{-1} a \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $

From this, we can deduce:

$ \sin^{-1} a \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] $

Consequently, the values for $ a $ must satisfy:

$ a \in \left[-\sin \frac{\pi}{4}, \sin \frac{\pi}{4}\right] $

This simplifies to:

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

Therefore, the condition for $ a $ is:

$ |a| \leq \frac{1}{\sqrt{2}} $

Given,

$ \begin{array}{c} f(x)=\sin ^{-1}\left(x^{2}-1\right)-3 \log _{3}\left(3^{x}-2\right) \\\\ -1 \leq x^{2}-1 \leq 1 \\\\ 0 \leq x^{2} \leq 2 \\\\ -\sqrt{2} < x < \sqrt{2} \end{array} $

$\therefore \sin ^{-1}\left(x^2-1\right)$ is not defined for $(-\infty,-\sqrt{2})$ and $(\sqrt{2}, 0)$

For $3 \log _3\left(3^x-2\right) \quad \Rightarrow \quad 3^x-2>0$

$ \begin{aligned} & 3^x>2 \\\\ & x>\log _3 2 \end{aligned} $

$\therefore 3 \log _3\left(3^x-2\right)$ is not defined for $x \leq \log _3(2)$

So, intervel $[-\sqrt{2}, \sqrt{2}]$ combine with $x>\log _3(2)$. The domain of $f(x)$ is $\log _3(2)

So, $f(x)$ is not defined for

$ \begin{aligned} & \left(-\infty, \log _3 2\right) \cup(\sqrt{2}, 0) \\\\ & a=\log _3 2, b=\sqrt{2} \\\\ & 3^a+b^2=2+2=4 \end{aligned} $

Given, $ \sin ^{-1}(4 x)-\cos ^{-1}(3 x)=\frac{\pi}{6} $

$ \Rightarrow \frac{\pi}{2}-\cos ^{-1} 4 x-\cos ^{-1} 3 x=\frac{\pi}{6} $

$ \Rightarrow \cos ^{-1} 4 x+\cos ^{-1} 3 x=\frac{\pi}{3} $

$ \Rightarrow 4 x \times 3 x-\sqrt{1-16 x^{2}} \sqrt{1-9 x^{2}}=\cos \frac{\pi}{3} $

$ \Rightarrow\left(12 x^{2}-\frac{1}{2}\right)^{2}=\left(1-16 x^{2}\right)\left(1-9 x^{2}\right) $

$ \Rightarrow 144 x^{4}+\frac{1}{4}-12 x^{2}=1-25 x^{2}+144 x^{4} $

$ 13 x^{2}=\frac{3}{4} \Rightarrow x^{2}=\frac{3}{4 \times 13} \Rightarrow x=\frac{\sqrt{3}}{2 \sqrt{13}} $

Given,

$ \sinh ^{-1}(-\sqrt{3})+\cosh ^{-1}(2)=K $

$ \log (-\sqrt{3}+\sqrt{3+1})+\log (2+\sqrt{3})=K $

$ \Rightarrow K=\log [(2-\sqrt{3})(2+\sqrt{3})] $

$ \Rightarrow K=\log 1=0 $

$ \cosh K=\cosh 0=\frac{e^{0}+e^{-0}}{2}=1 $

$ y=\cos ^{-1}\left(\frac{-2 x^{2}+6 x-4}{2 x^{2}-6 x+5}\right) $

$ \Rightarrow \frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(\frac{6 x-2 x^{2}-4}{2 x^{2}-6 x+5}\right)^{2}}} $

$ \begin{array}{l} \Rightarrow \frac{d y}{d x}=\frac{-1}{\sqrt{\frac{\left(2 x^{2}-6 x+5\right)^{2}-\left(-\left(2 x^{2}-6 x+4\right)^{2}\right.}{\left(2 x^{2}-6 x+5\right)}}} \\\\ \frac{\left(12 x^{2}-36 x+30-8 x^{3}+24 x^{2}-20 x\right)}{-\left(24 x^{2}-8 x^{3}-16 x-36 x+12 x^{2}+24\right.} \frac{\left(2 x^{2}-6 x+5\right)^{2}}{} \\\\ \Rightarrow \frac{d y}{d x}=\frac{-1\left(2 x^{2}-6 x+5\right)}{\sqrt{4 x^{2}-12 x+9}} . \\\\ 12 x^{2}-36 x+30-8 x^{3}+24 x^{2}-20 x \\\\ -24 x^{2}+8 x^{3}+16 x+36 x-12 x-24 \\\\ \Rightarrow \frac{d y}{d x}=\frac{-1\left(2 x^{2}-6 x+5\right)^{2}}{\sqrt{4 x^{2}-12 x+9}} \cdot \frac{(-4 x+6}{\left(2 x^{2}-6 x+5\right)^{2}} \\\\ \Rightarrow \frac{d y}{d x}=\frac{2(2 x-3)}{(2 x-3)\left(2 x^{2}-6 x+5\right)} \\\\ \Rightarrow \frac{d y}{d x}=\frac{2}{2 x^{2}-6 x+5} \end{array} $

Given, $ 2 \tan ^{-1} x=3 \sin ^{-1} x $ and $ x \neq 0 $

$ \begin{array}{l} \therefore \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\sin ^{-1}\left(3 x-4 x^{3}\right) \\\\ \Rightarrow \quad \frac{2 x}{1+x^{2}}=x\left(3-4 x^{2}\right) \\\\ \Rightarrow \quad 2=\left(1+x^{2}\right)\left(3-4 x^{2}\right) \quad[\because x \neq 0] \\\\ \Rightarrow \quad 4 x^{4}+x^{2}-1=0 \\\\ \Rightarrow \quad x^{2}=\frac{-1+\sqrt{(1)^{2}-4(4)(-1)}}{2(4)}\left[\because 0 < x^{2}\right] \\\\ \Rightarrow \quad 8 x^{2}=-1+\sqrt{17} \Rightarrow 8 x^{2}+1=\sqrt{17} \end{array} $

(a) $\sinh x$ is an odd function and its range is $R$ and domain is $R$.

(b) $\operatorname{sech} x$ is an even function and its range is ( 0,1 ] and domain is $R$.

(c) $\tanh x$ is an odd function and its range is $(-1,1)$ and domain is $R$.

(d) $\operatorname{cosec}^{-1} x$ is neither even nor odd function.

Hence, A-IV, B-III, C-V, D-II

We have,

$ \begin{aligned} & \cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\tan ^{-1} \frac{16}{63} \\\\ & =\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{16}{63} \\\\ & {\left[\because \cos ^{-1} \frac{3}{5}=\tan ^{-1} \frac{4}{3} \text { and } \sin ^{-1} \frac{5}{13}=\tan ^{-1} \frac{5}{12}\right]} \\\\ & =\tan ^{-1} \frac{63}{16}+\tan ^{-1} \frac{16}{63} \\\\ & =\cot ^{-1} \frac{16}{63}+\tan ^{-1} \frac{16}{63}=\frac{\pi}{2} \end{aligned} $

We have,

$ \begin{aligned} & \cosh ^{-1}\left(\frac{5}{3}\right)+\sinh ^{-1}\left(\frac{3}{4}\right)=k \\\\ & \ln \left(\frac{5}{3}+\sqrt{\left(\frac{5}{3}\right)^2-1}\right)+\ln \left[\left(\frac{3}{4}\right)+\sqrt{\left(\frac{3}{4}\right)^2+1}\right]=k \end{aligned} $

$\begin{aligned} & {\left[\begin{array}{r}\because \cos h^{-1} x=\ln \left(x+\sqrt{x^2-1}\right) \\\\ \text { and } \sin h^{-1} x=\ln \left(x+\sqrt{x^2+1}\right)\end{array}\right]} \\\\ & \ln \left(\frac{5}{3}+\frac{4}{3}\right)+\ln \left(\frac{3}{4}+\frac{5}{4}\right)=k \\\\ & \ln 3+\ln 2=k \\\\ & \ln 6=k \\\\ & e^k=6\end{aligned}$

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} $

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 $