Inverse Trigonometric Functions

Simplify the individual components

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

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

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

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

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

Rewrite the expression

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

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

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

3. Calculate the final value

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

We need to find the value of

$\cot^{-1}(\cot(-11)) + 10\sin\left(2\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) + 10\sin(2\tan^{-1}(2))$

Simplify $ \cot^{-1}(\cot(-11)) $

The principal value of $ \cot^{-1}x $ lies in the interval $ (0, \pi) $. We need to find an angle in this range that has the same cotangent as $ -11 $.

Now, $ -11 $ radians is less than $ 0 $. If we add multiples of $ \pi $ till it falls in $ (0, \pi) $, we get $ 4\pi - 11 $, which lies in $ (0, \pi) $.

Hence, $ \cot^{-1}(\cot(-11)) = 4\pi - 11 $.

Simplify $ 10\sin\left(2\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) $

Let $ \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \alpha $. Then, $ \cos \alpha = \frac{1}{\sqrt{2}} $. This gives $ \alpha = \frac{\pi}{4} $.

Now, $ 2\alpha = \frac{\pi}{2} $. $\sin\left(2\alpha\right) = \sin\left(\frac{\pi}{2}\right) = 1.$

Therefore, this term becomes $ 10 \times 1 = 10.$

Simplify $ 10\sin(2\tan^{-1}2) $

Let $ \tan^{-1}2 = \theta $, so $ \tan\theta = 2 $. Here, $ \theta \in \left(0, \frac{\pi}{2}\right) $.

We use the double angle formula: $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\times 2}{1 + 2^2} = \frac{4}{5}.$

Hence, $ 10\sin(2\tan^{-1}2) = 10 \times \frac{4}{5} = 8.$

Combine all terms

$\cot^{-1}(\cot(-11)) + 10\sin\left(2\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) + 10\sin(2\tan^{-1}(2))$

$= (4\pi - 11) + 10 + 8$

$= 4\pi + 7$

Final Answer: $ 4\pi + 7 $

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

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

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

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

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

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

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

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

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

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

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

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

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

We can express $ f(x) $ as:

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

This simplifies to:

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

Further simplification gives:

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

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

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

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

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

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

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

To find the total number of real solutions for the given equation, let's break down the problem step by step:

Define $\alpha$ as:

$ \alpha = \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $

We need to solve:

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

This can be rearranged as:

$ \theta + \alpha = \tan^{-1}(2 \tan \theta) $

Taking the tangent of both sides, we have:

$ \tan(\theta + \alpha) = 2 \tan \theta $

The tangent addition formula gives:

$ \frac{\tan \theta + \tan \alpha}{1 - \tan \alpha \tan \theta} = 2 \tan \theta \quad \ldots (1) $

Next, consider:

$ \sin 2\alpha = \frac{6 \tan \theta}{9 + \tan^2 \theta} = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} $

This leads to the equation:

$ 3 \tan \theta + 3 \tan \theta \tan^2 \alpha = 9 \tan \alpha + \tan \alpha \tan^2\theta $

Simplifying:

$ 3(\tan \theta - 3 \tan \alpha) = \tan \alpha \tan \theta (\tan \theta - 3 \tan \alpha) $

This implies:

$ \tan \theta = \frac{3}{\tan \alpha} \quad \text{or} \quad \tan \theta = 3 \tan \alpha $

Case I: $\tan \theta = 3 \tan \alpha$

From equation (1):

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

This gives:

$ \tan \theta = 0, \quad \frac{2}{3} = 1 - \frac{\tan^2 \theta}{3} \Rightarrow \tan \theta = 1, -1 $

Thus, we have:

$ \tan \theta = 0, -1, 1 \Rightarrow \theta = \frac{\pi}{4}, -\frac{\pi}{4}, 0 $

Case II: $\tan \theta = \frac{3}{\tan \alpha}$

From equation (1):

$ \frac{\tan \theta + \frac{3}{\tan \theta}}{-2} = 2 \tan \theta $

Simplifying this results in:

$ \tan \theta + \frac{3}{\tan \theta} = -4 \tan \theta $

This leads to no viable solutions (as denoted by "No Solution").

So, the real solutions for $\theta$ are $\frac{\pi}{4}$, $-\frac{\pi}{4}$, and $0$, resulting in a total of 3 real solutions.

$ \begin{aligned} & \text { } \tan ^{-1}(x)+\tan ^{-1}(2 x)=\tan ^{-1}(l) \\ & \tan ^{-1}\left(\frac{3 x}{1-2 x^2}\right)=\tan ^{-1}(l) \\ & 3 x=1-2 x^2 \\ & 2 x^2+3 x-1=0 \\ & \quad x=\frac{-3 \pm \sqrt{9+8}}{4} \\ & \quad x=\frac{-3 \pm \sqrt{17}}{4} \end{aligned} $

So, $\quad x=\frac{-3+\sqrt{17}}{4}$ will satisfy Number of solution is 1 .

$ \begin{aligned} &\text { A is false since }\\ &\begin{aligned} \sin ^{-1}(\log x) & +\cos ^{-1}(\log x) \\ & =\frac{\pi}{2} \text { only when }-1 \leq \log x \leq 1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { So, } \int \sqrt{x-3}\left(\sin ^{-1}(\log x)+\cos ^{-1}(\log x) d x\right. \\ & \qquad \neq \frac{\pi}{2} \times(x-3)^{\frac{3}{2}} \times \frac{2}{3}+C \\ & \sin ^{-1}(f(x))+\cos ^{-1}\left(f(x)=\frac{\pi}{2}\right. \text { when } \\ & |f(x)| \leq 1 \end{aligned}\\ &\text { So } A \text { is false and } B \text { is true. } \end{aligned} $

$ \text { } \begin{aligned} & \sin ^{-1}(-\cos 2)=-\sin ^{-1}\left(\sin \left(\frac{\pi}{2}-2\right)\right) \\ & \Rightarrow \quad \cos ^{-1}(\sin 3)=\cos ^{-1}\left(\cos \left(3-\frac{\pi}{2}\right)\right) \\ & \Rightarrow \quad \tan ^{-1}(\cot 5)=\tan ^{-1}\left(\tan \left(\frac{3 \pi}{2}-5\right)\right) \\ & \text { So, } \sin ^{-1}(-\cos 2)+\cos ^{-1}(\sin 3) \\ &+ \tan ^{-1}(\cot 5) \\ &=2-\frac{\pi}{2}+3-\frac{\pi}{2}+\frac{3 \pi}{2}-5 \\ &= \frac{3 \pi}{2}-\pi=\frac{\pi}{2} \end{aligned} $

$f(x)=\cos ^{-1}(2 x-5)-\sin ^{-1}(x-2)$

Domain of $f(x)$

$ \begin{aligned} & =|2 x-5| \leq 1 \cap|x-2| \leq 1 \\ & =-1 \leq 2 x-5 \leq 1 \cap-1 \leq x-2 \leq 1 \\ & =2 \leq x \leq 3 \cap 1 \leq x \leq 3, x \in[2,3] \\ f^{\prime}(x) & =\frac{-1 \times \hat{2}}{\sqrt{1-(2 x-5)^2}}-\frac{1 \times 1}{\sqrt{1-(x-2)^2}} \end{aligned} $

Now, $f^{\prime}(x)$ exist only

$ \begin{aligned} & \quad 1-(2 x-5)^2>0 \text { and } 1-(x-2)^2>0 \\ & \Rightarrow \quad x \in(2,3) \text {. } \\ & \text { Domain }=(2,3) \end{aligned} $

We have,

$ \begin{aligned} & \tan ^{-1}\left(x+\frac{\sqrt{2}}{x}\right)+\tan ^{-1}\left(x-\frac{\sqrt{2}}{x}\right) \\ & =\tan ^{-1} x \\ & x \neq 0 \\ & \text { Now, } \tan ^{-1}\left(\frac{x+\frac{\sqrt{2}}{x}+x-\frac{\sqrt{2}}{x}}{1-\left(x+\frac{\sqrt{2}}{x}\right)\left(x-\frac{\sqrt{2}}{x}\right)}\right) \\ & =\tan ^{-1} x \\ & \Rightarrow \quad \frac{2 x}{1-\left(x^2-\frac{2}{x^2}\right)}=x \\ & \Rightarrow \quad 2=1-x^2+\frac{2}{x^2} \\ & \quad x^2-\frac{2}{x^2}+1=0 \\ & \Rightarrow \quad x^4+x^2-2=0 \\ & \quad\left(x^2+2\right)\left(x^2-1\right)=0 \\ & \quad x^2=2 \Rightarrow x= \pm 1 \end{aligned} $

∴ Number of values of $x=2$

$ \begin{aligned} &\text { } y=\sec ^{-1} x\\ &\begin{aligned} & x=\sec y, \cos y=\frac{1}{x}, \sin y=\frac{\sqrt{x^2-1}}{x} \\ & 1=\sec y \tan y \frac{d y}{d x} \\ \Rightarrow & \frac{d y}{d x}=\cot y \cdot \cos y \\ \Rightarrow & \frac{d^2 y}{d x^2}=\left[\cot y \cdot(-\sin y)+\cos y\left(-\operatorname{cosec}^2 y\right)\right] \frac{d y}{d x} \\ = & -\cos y\left(1+\frac{1}{\sin ^2 y}\right) \cdot \cot y \cdot \cos y \\ = & -\cos ^3 y \frac{\left(1+\sin ^2 y\right)}{\sin ^3 y}=\frac{1}{x|x|} \frac{\left(1-2 x^2\right)}{\left(x^2-1\right)^{\frac{3}{2}}} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \tan ^{-1}(2 x-1) & +\tan ^{-1} 2 x \\ = & \tan ^{-1} 4 x-\tan ^{-1}(2 x+1) \\ \Rightarrow \tan ^{-1}(2 x-1) & +\tan ^{-1}(2 x+1) \\ = & \tan ^{-1} 4 x-\tan ^{-1} 2 x \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \tan ^{-1}\left(\frac{2 x-1+2 x+1}{1-(2 x-1)(2 x+1)}\right) \\ & =\tan ^{-1}\left(\frac{4 x-2 x}{1+(4 x)(2 x)}\right) \\ & \Rightarrow \tan ^{-1}\left(\frac{4 x}{1-4 x^2+1}\right)=\tan ^{-1}\left(\frac{2 x}{1+8 x^2}\right) \\ & \therefore \frac{4 x}{2-4 x^2}=\frac{2 x}{1+8 x^2} \\ & \quad x=0 \\ & \quad 1-x^2=1+8 x^2 \\ & \quad x^2=0 \Rightarrow x=0 \end{aligned}\\ &\text { ∴ Only one solution. } \end{aligned} $

$ \begin{aligned} & \sinh ^{-1} x=\cosh ^{-1} y=\log (1+\sqrt{2}) \\ & \sinh ^{-1} x=\log (1+\sqrt{2}) \\ & \log \left(x+\sqrt{x^2}+1\right)=\log (1+\sqrt{2}) \\ & x=1 \\ & \text { And } \cosh ^{-1} y=\log (1+\sqrt{2}) \\ & \Rightarrow \log \left(y+\sqrt{y^2-1}\right)=\log (1+\sqrt{2}) \\ & \therefore \quad y=\sqrt{2} \end{aligned} $

Now, $\tan ^{-1}(x+y)$

$ \begin{aligned} & =\tan ^{-1}(1+\sqrt{2}) \\ & =67.5^{\circ} \end{aligned} $

By LHS, $\tan ^{-1}\left(\sqrt{\frac{x(x+y+z)}{y z}}\right)+\tan ^{-1}\left(\sqrt{\frac{y(x+y+z)}{x z}}\right)$$ +\tan ^{-1}\left(\sqrt{\frac{z(x+y+z)}{x y}}\right) $

Let $a=\sqrt{\frac{x(x+y+z)}{y z}}, b=\sqrt{\frac{y(x+y+z)}{x z}}$ and $z=\sqrt{\frac{z(x+y+z)}{x y}}$

Here, $a+b+c=\frac{(x+y+z)^{1 / 2}(x+y+z)}{\sqrt{x y z}}=\frac{(x+y+z)^{3 / 2}}{\sqrt{x y z}}$

And $a b c=\frac{(x+y+z)^{3 / 2}}{\sqrt{x y z}}$

Since, $a+b+c=a b c$

$ \therefore \quad \text { LHS }=\pi $

$\Rightarrow$ Assertion is true and since, $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$.

if $x y<1$

$\Rightarrow$ Reason is false.

$ \begin{aligned} &\text { We know that }\\ &\begin{aligned} & \sinh ^{-1} x=\log \left(x+\sqrt{x^2+1}\right) \\ &\therefore \sinh ^{-1} 2=\log (2+\sqrt{4+1})=\log (2+\sqrt{5}) \\ & \cosh ^{-1} x=\log \left(x+\sqrt{x^2-1}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \therefore \cosh ^{-1} \sqrt{6}=\log (\sqrt{6}+\sqrt{6-1})=\log (\sqrt{6}+\sqrt{5}) \\ & \therefore \sinh ^{-1} 2+\cosh ^{-1} \sqrt{6} \\ & =\log (2+\sqrt{5})+\log (\sqrt{6}+\sqrt{5}) \\ & =\log [(2+\sqrt{5})(\sqrt{6}+\sqrt{5})] \\ & e^{\left(\sinh ^{-1} \sqrt{2}+\cosh ^{-1} \sqrt{6}\right)}=e^{\log ((2+\sqrt{5})(\sqrt{6}+\sqrt{5})]} \\ & =(2+\sqrt{5})(\sqrt{6}+\sqrt{5})=2 \sqrt{6}+2 \sqrt{5}+\sqrt{30}+5 \\ & =5+2 \sqrt{5}+2 \sqrt{6}+\sqrt{30}=5+2 \sqrt{5}+2 \sqrt{6}+\sqrt{5} \sqrt{6} \\ & =5+(2+\sqrt{6}) \sqrt{5}+2 \sqrt{6}=a+(b+\sqrt{c}) \sqrt{a}+b \sqrt{c} \\ & \therefore a=5, b=2, c=6 \\ & \therefore \quad a+b+c=5+2+6=13 \end{aligned} $

Given,

$ \begin{aligned} & \frac{d}{d x}\left(\tan ^{-1}\left(\frac{1+x}{1-x}\right)\right)=\frac{d}{d x}\left(\tan ^{-1} x\right) \\ & \text { By LHS, } \frac{d}{d x}\left(\tan ^{-1}\left(\frac{1+x}{1-x}\right)\right) \\ & =\frac{1}{1+\left(\frac{1+x}{1-x}\right)^2} \frac{d}{d x}\left(\frac{1+x}{1-x}\right) \\ & =\frac{(1-x)^2}{(1-x)^2+(1+x)^2}\left[\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^2}\right] \\ & =\frac{(1-x)^2}{2\left(1+x^2\right)}\left[\frac{1-x+1+x}{(1-x)^2}\right]=\frac{1}{(1+x)^2} \end{aligned} $

And by RHS, $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$

$\therefore$ Assertion is true.

For reason ( R )

when $x<1$, let $x=\tan \theta$

$ \begin{array}{ll} \Rightarrow & \theta=\tan ^{-1} x \\ \therefore & \tan ^{-1}\left(\frac{1+x}{1-x}\right)=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \\ & =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right)=\frac{\pi}{4}+\theta \\ & =\frac{\pi}{4}+\tan ^{-1} x \end{array} $

When $x>1$, then $\tan ^{-1} x>1$

$ \therefore \quad \frac{1+x}{1-x}<0 $

so, $\tan ^{-1}\left(\frac{1+x}{1-x}\right)$ lies in $\left(-\frac{\pi}{2}, 0\right)$

$ \therefore \tan ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{-3 \pi}{4}+\tan ^{-1} x $

$\therefore$ Reason is also true.

We have, $y=\left(\sin ^{-1} x\right)^2$

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=2\left(\sin ^{-1} x\right) \frac{1}{\sqrt{1-x^2}} \\ & \Rightarrow \quad \sqrt{1-x^2} \frac{d y}{d x}=2 \sin ^{-1} x \\ & \Rightarrow \quad\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=4\left(\sin ^{-1} x\right)^2 \\ & \Rightarrow \quad\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=4 y \text { (Using Eq. } \end{aligned} $

Again, differentiating w.r.t $x$, we get

$ \begin{aligned} & (-2 x)\left(\frac{d y}{d x}\right)^2+\left(1-x^2\right)\left(2 \frac{d y}{d x}\right) \frac{d^2 y}{d x^2}=4 \frac{d y}{d x} \\ \Rightarrow & (-2 x) \frac{d y}{d x}+2\left(1-x^2\right) \frac{d^2 y}{d x^2}=4 \\ \Rightarrow & \left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=2 \end{aligned} $

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

The domain of $\sin ^{-1}\left(\sqrt{x^2+x+1}\right)$ is $[-1,1]$

$ \therefore \quad-1 \leq \sqrt{x^2+x+1} \leq 1 $

Since $\sqrt{x^2+x+1}$ is always,

Let $g(x)=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$

The minimum value of $g(x)$ is $3 / 4$, which occurs at $x=-\frac{1}{2}$

$ \begin{array}{ll} \therefore & x^2+x+1 \geq \frac{3}{4} \\ \Rightarrow & \sqrt{x^2+x+1} \geq \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \\ \Rightarrow & \frac{\sqrt{3}}{2} \leq \sqrt{x^2+x+1} \leq 1 \\ \Rightarrow & \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right) \leq \sin ^{-1}\left(\sqrt{x^2+x+1}\right) \\ \Rightarrow & \frac{\pi}{3} \leq f(x) \leq \frac{\pi}{2} \quad \leq \sin ^{-1} \quad \end{array} $

Thus, the range of $f(x)=\sin ^{-1}\left(\sqrt{x^2+x+1}\right)$ is $\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$

$ \begin{aligned} & \text { } \begin{aligned} \tan & \left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{6}{41}\right) \\ & =\tan ^{-1}\left(\frac{\frac{3}{5}+\frac{6}{41}}{1-\frac{3}{5} \cdot \frac{6}{41}}\right)=\tan ^{-1}\left(\frac{\frac{153}{205}}{\frac{187}{205}}\right) \\ & =\tan ^{-1}\left(\frac{153}{187}\right) \\ \text { Now, } & \tan { }^{-1}\left(\frac{153}{187}\right)+\tan ^{-1}\left(\frac{9}{191}\right) \\ & =\tan ^{-1}\left(\frac{\frac{153}{187}+\frac{9}{191}}{1-\frac{153}{187} \cdot \frac{9}{191}}\right) \\ & =\tan ^{-1}\left(\frac{\frac{29223+1683}{35717}}{1-\frac{1377}{35717}}\right) \\ & =\tan ^{-1}\left(\frac{30906}{34340}\right)=\tan ^{-1}\left(\frac{9}{10}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } 2 \tanh ^{-1} x=\sinh ^{-1}\left(\frac{4}{3}\right) \\ & \begin{array}{l} \tanh ^{-1}(x)=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \\ \Rightarrow \quad 2 \tanh ^{-1}(x)=\ln \left(\frac{1+x}{1-x}\right) \\ \text { So, } \ln \left(\frac{1+x}{1-x}\right)=\sinh ^{-1}\left(\frac{4}{3}\right) \\ =\ln \left(\frac{4}{3}+\sqrt{\left(\frac{4}{3}\right)^2+1}\right) \\ {\left[\because \sinh ^{-1}(y)=\ln \left(y+\sqrt{y^2+1}\right)\right]} \end{array} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\ln \left(\frac{4}{3}+\sqrt{\frac{16}{9}+1}\right) \\ & =\ln \left(\frac{4}{3}+\frac{5}{3}\right) \Rightarrow \ln \frac{9}{3}=\ln (3) \\ & =\frac{1+x}{1-x}=3 \Rightarrow 1+x=3-3 x \\ & =4 x=2 \Rightarrow x=\frac{1}{2} \end{aligned}\\ &\text { Now, } \cosh ^{-1}\left(\frac{1}{x}\right)=\cosh ^{-1}(4\\ &\begin{aligned} & =\ln (2+\sqrt{4-1}) \\ & \quad\left[\because \cosh ^{-1}(z)=\ln \left(z+\sqrt{z^2-1}\right)\right] \\ & =\ln (2+\sqrt{3}) \end{aligned} \end{aligned} $

Given, $f(x)=\sqrt{\cos ^{-1} \sqrt{1-x^2}}$

Let $x=\sin \theta$.

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

$ \Rightarrow \quad \sqrt{\cos ^2 \theta}=|\cos \theta| $

Since, $\theta \in\left[0, \frac{\pi}{2}\right]$, so $\cos \theta \geq 0$

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

So, $f(x)=\sqrt{\cos ^{-1}(\cos \theta)}$

$ \begin{aligned} \Rightarrow \quad \sqrt{\theta} & =\sqrt{\sin ^{-1}(x)} \\ f^{\prime}(x) & =\frac{d}{d x} \sqrt{\sin ^{-1}(x)} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad & \frac{1}{\sqrt{1-x^2}} \cdot \frac{1}{2 \sqrt{\sin ^{-1}(x)}} \\ & f^{\prime}(x)=\frac{1}{\sqrt{1-\left(\frac{1}{2}\right)^2}} \cdot \frac{1}{2 \cdot \sqrt{\sin ^{-1} \frac{1}{2}}} \\ & =\frac{1}{\frac{\sqrt{3}}{2}} \cdot \frac{1}{2 \cdot \sqrt{\frac{\pi}{6}}} \end{aligned} $

$ \Rightarrow \quad \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{6}}{2 \sqrt{\pi}} \Rightarrow \frac{\sqrt{2}}{\sqrt{\pi}}=\sqrt{\frac{2}{\pi}} $

$ \therefore \quad f^{\prime}(x)=\sqrt{\frac{2}{\pi}} $

$ \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) $