Mix Examples-ITF Questions in English

(B) Given equation: $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$
$\Rightarrow 2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=n \pi, n \in \mathbb{Z}$
$\Rightarrow \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=\frac{n \pi}{2}$
Since the range of $\cos ^{-1} \theta$ is $[0, \pi]$,we have $\cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\} \in \{0, \frac{\pi}{2}, \pi\}$.
Case $1$: $\cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=0 \Rightarrow \cot \left(2 \tan ^{-1} x\right)=1 \Rightarrow 2 \tan ^{-1} x = \frac{\pi}{4} + m\pi \Rightarrow \tan ^{-1} x = \frac{\pi}{8} + \frac{m\pi}{2}$.
For $x \ge 1$,$\tan ^{-1} x \in [\frac{\pi}{4}, \frac{\pi}{2})$. The only value is $\tan ^{-1} x = \frac{\pi}{4} \Rightarrow x = 1$.
Case $2$: $\cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=\frac{\pi}{2} \Rightarrow \cot \left(2 \tan ^{-1} x\right)=0 \Rightarrow 2 \tan ^{-1} x = \frac{\pi}{2} + m\pi \Rightarrow \tan ^{-1} x = \frac{\pi}{4} + \frac{m\pi}{2}$.
For $x \ge 1$,$\tan ^{-1} x = \frac{\pi}{4} \Rightarrow x = 1$ (already counted).
Case $3$: $\cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=\pi \Rightarrow \cot \left(2 \tan ^{-1} x\right)=-1 \Rightarrow 2 \tan ^{-1} x = \frac{3\pi}{4} + m\pi \Rightarrow \tan ^{-1} x = \frac{3\pi}{8} + \frac{m\pi}{2}$.
For $x \ge 1$,$\tan ^{-1} x = \frac{3\pi}{8} \Rightarrow x = \tan \frac{3\pi}{8} = \sqrt{2} + 1$.
Thus,the roots $\ge 1$ are $1$ and $\sqrt{2} + 1$.
The number of such roots is $2$.

(A) Let $a = \cos^{-1} x$ and $b^2 = (\sin^{-1} y)^2$. The given equations are $a + b^2 = \frac{n \pi^2}{4}$ and $a \cdot b^2 = \frac{\pi^4}{16}$.
These are roots of the quadratic equation $t^2 - (\frac{n \pi^2}{4})t + \frac{\pi^4}{16} = 0$.
For real roots,the discriminant $D \ge 0$,so $(\frac{n \pi^2}{4})^2 - 4(\frac{\pi^4}{16}) \ge 0$.
$\frac{n^2 \pi^4}{16} - \frac{\pi^4}{4} \ge 0 \Rightarrow n^2 \ge 4 \Rightarrow n \ge 2$ (since $n \in Z$ and $n > 0$ for real $x, y$).
For the least value $n = 2$,the quadratic becomes $t^2 - \frac{2 \pi^2}{4}t + \frac{\pi^4}{16} = 0$,which is $(t - \frac{\pi^2}{4})^2 = 0$.
Thus,$a = \frac{\pi^2}{4}$ and $b^2 = \frac{\pi^2}{4}$.
Since $a = \cos^{-1} x = \frac{\pi^2}{4}$,we have $x = \cos(\frac{\pi^2}{4})$.
Since $b^2 = (\sin^{-1} y)^2 = \frac{\pi^2}{4}$,we have $\sin^{-1} y = \pm \frac{\pi}{2}$,so $y = \sin(\pm \frac{\pi}{2}) = \pm 1$.
The solution is $(\cos(\frac{\pi^2}{4}), \pm 1)$.

(D) Given $2 \tan^{-1} x = 3 \sin^{-1} x$.
Using the identity $2 \tan^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$ and $3 \sin^{-1} x = \sin^{-1} (3x - 4x^3)$,we have:
$\sin^{-1} \left( \frac{2x}{1+x^2} \right) = \sin^{-1} (3x - 4x^3)$
$\Rightarrow \frac{2x}{1+x^2} = 3x - 4x^3$
Since $x \neq 0$,we can divide by $x$:
$\frac{2}{1+x^2} = 3 - 4x^2$
$2 = (3 - 4x^2)(1 + x^2)$
$2 = 3 + 3x^2 - 4x^2 - 4x^4$
$4x^4 + x^2 - 1 = 0$
Using the quadratic formula for $x^2$:
$x^2 = \frac{-1 \pm \sqrt{1^2 - 4(4)(-1)}}{2(4)} = \frac{-1 \pm \sqrt{17}}{8}$
Since $x^2 > 0$,we take $x^2 = \frac{\sqrt{17} - 1}{8}$.
Then $8x^2 = \sqrt{17} - 1$,which implies $8x^2 + 1 = \sqrt{17}$.

(D) Given: $\sinh ^{-1}(-\sqrt{3})+\cosh ^{-1}(2)=K$
Let $x=\sinh ^{-1}(-\sqrt{3})$ and $y=\cosh ^{-1}(2)$.
Then $x+y=K$.
From the definitions,$\sinh x = -\sqrt{3}$ and $\cosh y = 2$.
Using the identity $\cosh^2 x - \sinh^2 x = 1$,we have $\cosh^2 x = 1 + (-\sqrt{3})^2 = 1 + 3 = 4$.
Since $\cosh x \geq 1$,we get $\cosh x = 2$.
Using the identity $\cosh^2 y - \sinh^2 y = 1$,we have $2^2 - \sinh^2 y = 1$,so $\sinh^2 y = 3$.
Since $\cosh^{-1}(2)$ is positive,$\sinh y = \sqrt{3}$.
Now,$\cosh K = \cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y$.
Substituting the values: $\cosh K = (2)(2) + (-\sqrt{3})(\sqrt{3}) = 4 - 3 = 1$.

(B) Given equation: $\operatorname{Tan}^{-1}(2x-1) + \operatorname{Tan}^{-1}(2x+1) = \operatorname{Tan}^{-1}(4x) - \operatorname{Tan}^{-1}(2x)$.
Using the formula $\operatorname{Tan}^{-1} A + \operatorname{Tan}^{-1} B = \operatorname{Tan}^{-1} \left( \frac{A+B}{1-AB} \right)$:
$\operatorname{Tan}^{-1} \left( \frac{(2x-1) + (2x+1)}{1 - (2x-1)(2x+1)} \right) = \operatorname{Tan}^{-1} \left( \frac{4x - 2x}{1 + (4x)(2x)} \right)$.
$\operatorname{Tan}^{-1} \left( \frac{4x}{1 - (4x^2 - 1)} \right) = \operatorname{Tan}^{-1} \left( \frac{2x}{1 + 8x^2} \right)$.
$\operatorname{Tan}^{-1} \left( \frac{4x}{2 - 4x^2} \right) = \operatorname{Tan}^{-1} \left( \frac{2x}{1 + 8x^2} \right)$.
Equating the arguments: $\frac{4x}{2(1 - 2x^2)} = \frac{2x}{1 + 8x^2}$.
$\frac{2x}{1 - 2x^2} = \frac{2x}{1 + 8x^2}$.
This implies $2x = 0$ or $\frac{1}{1 - 2x^2} = \frac{1}{1 + 8x^2}$.
Case $1$: $2x = 0 \implies x = 0$. Since $0 \leq 0 < \frac{3}{4}$,$x = 0$ is a valid solution.
Case $2$: $1 - 2x^2 = 1 + 8x^2 \implies 10x^2 = 0 \implies x = 0$.
Thus,the only solution is $x = 0$. The number of values is $1$.

(C) Let the given equation be $\operatorname{Tan}^{-1}\left(x+\frac{\sqrt{2}}{x}\right)+\operatorname{Tan}^{-1}\left(x-\frac{\sqrt{2}}{x}\right)=\operatorname{Tan}^{-1}(x)$.
Applying the formula $\operatorname{Tan}^{-1}(A) + \operatorname{Tan}^{-1}(B) = \operatorname{Tan}^{-1}\left(\frac{A+B}{1-AB}\right)$,we get:
$\operatorname{Tan}^{-1}\left(\frac{x+\frac{\sqrt{2}}{x} + x-\frac{\sqrt{2}}{x}}{1-(x+\frac{\sqrt{2}}{x})(x-\frac{\sqrt{2}}{x})}\right) = \operatorname{Tan}^{-1}(x)$.
Simplifying the expression inside the $\operatorname{Tan}^{-1}$ function:
$\frac{2x}{1-(x^2 - \frac{2}{x^2})} = x$.
$\frac{2x}{1-x^2 + \frac{2}{x^2}} = x$.
Assuming $x \neq 0$,we can divide by $x$:
$\frac{2}{1-x^2 + \frac{2}{x^2}} = 1$.
$2 = 1 - x^2 + \frac{2}{x^2}$.
$x^2 - \frac{2}{x^2} + 1 = 0$.
Let $t = x^2$,then $t - \frac{2}{t} + 1 = 0 \implies t^2 + t - 2 = 0$.
$(t+2)(t-1) = 0$.
Since $t = x^2$,$t$ must be positive,so $t = 1$,which means $x^2 = 1$,so $x = \pm 1$.
Checking $x = 1$: $\operatorname{Tan}^{-1}(1+\sqrt{2}) + \operatorname{Tan}^{-1}(1-\sqrt{2}) = \operatorname{Tan}^{-1}(1)$. This is true as $\operatorname{Tan}^{-1}(1+\sqrt{2}) + \operatorname{Tan}^{-1}(1-\sqrt{2}) = \frac{3\pi}{8} + \frac{\pi}{8} = \frac{\pi}{2}$ is not correct,but using $\operatorname{Tan}^{-1}(A) + \operatorname{Tan}^{-1}(B) = \operatorname{Tan}^{-1}(\frac{A+B}{1-AB})$ holds for $AB < 1$. Here $AB = x^2 - 2/x^2 = 1 - 2 = -1 < 1$. Thus,$x=1$ and $x=-1$ are solutions. There are $2$ values.

(B) Given the equation $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} 2x = \frac{\pi}{4}$.
Using the formula $\operatorname{Tan}^{-1} A + \operatorname{Tan}^{-1} B = \operatorname{Tan}^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$\operatorname{Tan}^{-1} \left( \frac{x+2x}{1-x(2x)} \right) = \frac{\pi}{4}$.
Taking $\tan$ on both sides:
$\frac{3x}{1-2x^2} = \tan \left( \frac{\pi}{4} \right) = 1$.
This simplifies to $3x = 1 - 2x^2$,or $2x^2 + 3x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$.
Since $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} 2x = \frac{\pi}{4}$ and $\frac{\pi}{4} > 0$,we must have $x > 0$ (as $x$ and $2x$ have the same sign).
Checking the values: $x = \frac{-3 + \sqrt{17}}{4} \approx \frac{-3 + 4.12}{4} = 0.28 > 0$ (Valid).
$x = \frac{-3 - \sqrt{17}}{4} < 0$ (Invalid).
Thus,there is exactly $1$ real solution.

(D) Let the given expression be $S = \sin \left(\tan ^{-1} \frac{4}{5}+\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}-\tan ^{-1} \frac{1}{7}\right)$.
Rearranging the terms,we get $S = \sin \left(\left(\tan ^{-1} \frac{4}{5}-\tan ^{-1} \frac{1}{7}\right) + \left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}\right)\right)$.
Let $p = \tan ^{-1} \frac{4}{5}-\tan ^{-1} \frac{1}{7}$. Using the formula $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \left(\frac{x-y}{1+xy}\right)$,we have $p = \tan ^{-1} \left(\frac{\frac{4}{5}-\frac{1}{7}}{1+\frac{4}{5} \cdot \frac{1}{7}}\right) = \tan ^{-1} \left(\frac{\frac{28-5}{35}}{\frac{35+4}{35}}\right) = \tan ^{-1} \left(\frac{23}{39}\right)$.
Let $q = \tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}$. Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left(\frac{x+y}{1-xy}\right)$,we have $q = \tan ^{-1} \left(\frac{\frac{4}{3}+\frac{1}{9}}{1-\frac{4}{3} \cdot \frac{1}{9}}\right) = \tan ^{-1} \left(\frac{\frac{12+1}{9}}{\frac{27-4}{27}}\right) = \tan ^{-1} \left(\frac{13}{9} \cdot \frac{27}{23}\right) = \tan ^{-1} \left(\frac{39}{23}\right)$.
Since $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$,we have $q = \cot ^{-1} \left(\frac{23}{39}\right) = \frac{\pi}{2} - \tan ^{-1} \left(\frac{23}{39}\right)$.
Thus,$p+q = \tan ^{-1} \left(\frac{23}{39}\right) + \frac{\pi}{2} - \tan ^{-1} \left(\frac{23}{39}\right) = \frac{\pi}{2}$.
Therefore,$S = \sin \left(\frac{\pi}{2}\right) = 1$.

(A) $(I)$ We have $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(\tan ^{-1} x\right)\right)\right)$.
Let $\tan ^{-1} x = \theta$,then $\tan \theta = x$. Thus,$\cos \theta = \frac{1}{\sqrt{1+x^2}}$.
So,$f(x) = \sin \left(\cot ^{-1} \left(\frac{1}{\sqrt{1+x^2}}\right)\right)$.
Let $\cot ^{-1} \left(\frac{1}{\sqrt{1+x^2}}\right) = \alpha$,then $\cot \alpha = \frac{1}{\sqrt{1+x^2}}$,which implies $\tan \alpha = \sqrt{1+x^2}$.
Then $\sin \alpha = \frac{\tan \alpha}{\sqrt{1+\tan^2 \alpha}} = \frac{\sqrt{1+x^2}}{\sqrt{1+1+x^2}} = \sqrt{\frac{1+x^2}{2+x^2}}$.
Therefore,$f(0) = \sqrt{\frac{1+0}{2+0}} = \frac{1}{\sqrt{2}}$.
Since $\frac{1}{\sqrt{2}} \neq \frac{1}{2}$,Statement $I$ is false.
$(II)$ We use the formula $4 \tan ^{-1} \frac{1}{5} = \tan ^{-1} \frac{120}{119}$.
Then $\sin \left(\tan ^{-1} \frac{120}{119} - \tan ^{-1} \frac{1}{239}\right) = \sin \left(\tan ^{-1} \left(\frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \times \frac{1}{239}}\right)\right) = \sin \left(\tan ^{-1} \left(\frac{28680 - 119}{28441 + 120}\right)\right) = \sin \left(\tan ^{-1} \frac{28561}{28561}\right) = \sin \left(\tan ^{-1} 1\right) = \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.
Since $\frac{1}{\sqrt{2}} \neq 1$,Statement $II$ is false.

(B) Given equation: $\tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8}$
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} \left( \frac{a+b}{1-ab} \right)$,we get:
$\tan ^{-1} \left( \frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \cdot \frac{1}{8}} \right) + \frac{1}{2} \sec ^{-1} x = \frac{\pi}{8}$
$\tan ^{-1} \left( \frac{\frac{13}{40}}{\frac{39}{40}} \right) + \frac{1}{2} \sec ^{-1} x = \frac{\pi}{8}$
$\tan ^{-1} \left( \frac{1}{3} \right) + \frac{1}{2} \sec ^{-1} x = \frac{\pi}{8}$
Multiply by $2$: $2 \tan ^{-1} \left( \frac{1}{3} \right) + \sec ^{-1} x = \frac{\pi}{4}$
Using $2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$:
$\tan ^{-1} \left( \frac{2/3}{1-1/9} \right) + \sec ^{-1} x = \frac{\pi}{4}$
$\tan ^{-1} \left( \frac{3}{4} \right) + \sec ^{-1} x = \frac{\pi}{4}$
Since $\sec ^{-1} x = \tan ^{-1} \sqrt{x^2-1}$,we have:
$\tan ^{-1} \left( \frac{3}{4} \right) + \tan ^{-1} \sqrt{x^2-1} = \frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{\frac{3}{4} + \sqrt{x^2-1}}{1 - \frac{3}{4} \sqrt{x^2-1}} = \tan \left( \frac{\pi}{4} \right) = 1$
$3 + 4 \sqrt{x^2-1} = 4 - 3 \sqrt{x^2-1}$
$7 \sqrt{x^2-1} = 1$
$\sqrt{x^2-1} = \frac{1}{7}$
$x^2 - 1 = \frac{1}{49}$
$x^2 = 1 + \frac{1}{49} = \frac{50}{49}$

(B) Given,$\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$.
We know that $\sec^{-1}(\frac{1}{x}) = \cos^{-1}(x)$.
So,$\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}$.
Also,$\tan^{-1}(2) = \sin^{-1}(\frac{2}{\sqrt{1+2^2}}) = \sin^{-1}(\frac{2}{\sqrt{5}})$.
Therefore,$\sin(\sin^{-1}(\frac{2}{\sqrt{5}})) = \frac{2}{\sqrt{5}}$.
Equating both sides: $\frac{\sqrt{1-x^2}}{x} = \frac{2}{\sqrt{5}}$.
Squaring both sides: $\frac{1-x^2}{x^2} = \frac{4}{5}$.
$5 - 5x^2 = 4x^2$.
$9x^2 = 5$.
$x^2 = \frac{5}{9}$.
Since $x>0$,$x = \frac{\sqrt{5}}{3}$.

(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$
Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we have:
$\sin ^{-1}(1-x) = \frac{\pi}{2} - \sin ^{-1} x - \sin ^{-1} x = \frac{\pi}{2} - 2 \sin ^{-1} x$
Taking $\sin$ on both sides:
$1-x = \sin(\frac{\pi}{2} - 2 \sin ^{-1} x) = \cos(2 \sin ^{-1} x)$
Using the identity $\cos(2\theta) = 1 - 2\sin^2\theta$,where $\theta = \sin ^{-1} x$:
$1-x = 1 - 2(\sin(\sin ^{-1} x))^2$
$1-x = 1 - 2x^2$
$2x^2 - x = 0$
$x(2x - 1) = 0$
Thus,$x = 0$ or $x = \frac{1}{2}$.
Both values satisfy the domain of the inverse trigonometric functions.
Therefore,$x \in \{0, \frac{1}{2}\}$.

(C) Let $\theta = \tan ^{-1} 2$,then $\tan \theta = 2$. We know that $\sec ^2 \theta = 1 + \tan ^2 \theta$.
Thus,$\sec ^2(\tan ^{-1} 2) = 1 + (2)^2 = 1 + 4 = 5$.
Let $\phi = \cot ^{-1} 3$,then $\cot \phi = 3$. We know that $\operatorname{cosec}^2 \phi = 1 + \cot ^2 \phi$.
Thus,$\operatorname{cosec}^2(\cot ^{-1} 3) = 1 + (3)^2 = 1 + 9 = 10$.
Adding these values,we get $5 + 10 = 15$.

(B) Given equation: $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$.
Since $\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$ and $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$,we have $\sin ^{-1} x-\cos ^{-1} 2 x=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$.
Thus,$\sin ^{-1} x=\frac{\pi}{6}+\cos ^{-1} 2 x$.
Taking sine on both sides: $x=\sin\left(\frac{\pi}{6}+\cos ^{-1} 2 x\right) = \sin\left(\frac{\pi}{6}\right)\cos\left(\cos ^{-1} 2 x\right) + \cos\left(\frac{\pi}{6}\right)\sin\left(\cos ^{-1} 2 x\right)$.
$x = \frac{1}{2}(2x) + \frac{\sqrt{3}}{2}\sqrt{1-(2x)^2} = x + \frac{\sqrt{3}}{2}\sqrt{1-4x^2}$.
This implies $\frac{\sqrt{3}}{2}\sqrt{1-4x^2} = 0$,so $1-4x^2=0$,which gives $x = \frac{1}{2}$ (as $x = -\frac{1}{2}$ does not satisfy the original equation).
Now,calculate $\tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)$ for $x=\frac{1}{2}$:
$\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1/2}{1/2+1}\right) = \tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)$.
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1}\left(\frac{a+b}{1-ab}\right)$:
$\tan ^{-1}\left(\frac{1/2+1/3}{1-(1/2)(1/3)}\right) = \tan ^{-1}\left(\frac{5/6}{5/6}\right) = \tan ^{-1}(1) = \frac{\pi}{4}$.

(B) Given $f(x)=ax+b$ is an onto function from $[-1,1]$ to $[0,2]$. Since $a>0$,$f(x)$ is an increasing function. Thus,$f(-1)=0$ and $f(1)=2$.
$-a+b=0 \implies a=b$.
$a+b=2 \implies 2a=2 \implies a=1, b=1$.
So,$f(x)=x+1$.
Now,evaluate the expression:
$\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{1}{5} = \tan ^{-1} \left( \frac{\frac{1}{8}+\frac{1}{5}}{1-\frac{1}{8} \times \frac{1}{5}} \right) = \tan ^{-1} \left( \frac{13/40}{39/40} \right) = \tan ^{-1} \frac{1}{3}$.
Then,$\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3} = \tan ^{-1} \left( \frac{\frac{1}{7}+\frac{1}{3}}{1-\frac{1}{7} \times \frac{1}{3}} \right) = \tan ^{-1} \left( \frac{10/21}{20/21} \right) = \tan ^{-1} \frac{1}{2}$.
Finally,$\cot \left( \tan ^{-1} \frac{1}{2} \right) = \cot \left( \cot ^{-1} 2 \right) = 2$.
Since $f(1)=1+1=2$,the result is $f(1)$.

(B) Let $\theta = \sin ^{-1}\left(\frac{12}{13}\right)$. Then $\sin \theta = \frac{12}{13}$. Using the Pythagorean theorem,the base is $\sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$. Thus,$\tan \theta = \frac{12}{5}$,so $\theta = \tan ^{-1}\left(\frac{12}{5}\right)$.
Let $\phi = \cos ^{-1}\left(\frac{4}{5}\right)$. Then $\cos \phi = \frac{4}{5}$. The perpendicular is $\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$. Thus,$\tan \phi = \frac{3}{4}$,so $\phi = \tan ^{-1}\left(\frac{3}{4}\right)$.
Substituting these into the expression:
$\tan ^{-1}\left(\frac{12}{5}\right) + \tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{63}{16}\right)$
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \pi + \tan ^{-1}\left(\frac{A+B}{1-AB}\right)$ for $AB > 1$:
$A = \frac{12}{5}, B = \frac{3}{4} \Rightarrow AB = \frac{36}{20} = 1.8 > 1$.
So,$\tan ^{-1}\left(\frac{12}{5}\right) + \tan ^{-1}\left(\frac{3}{4}\right) = \pi + \tan ^{-1}\left(\frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \times \frac{3}{4}}\right) = \pi + \tan ^{-1}\left(\frac{\frac{48+15}{20}}{1 - \frac{36}{20}}\right) = \pi + \tan ^{-1}\left(\frac{\frac{63}{20}}{-\frac{16}{20}}\right) = \pi + \tan ^{-1}\left(-\frac{63}{16}\right) = \pi - \tan ^{-1}\left(\frac{63}{16}\right)$.
Adding the final term:
$\pi - \tan ^{-1}\left(\frac{63}{16}\right) + \tan ^{-1}\left(\frac{63}{16}\right) = \pi$.

(B) Let $\theta = \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)$,then $\cos 2 \theta = \frac{a}{b}$.
The expression is $\tan \left(\frac{\pi}{4} + \theta\right) + \tan \left(\frac{\pi}{4} - \theta\right)$.
Using the identity $\tan(A+B) + \tan(A-B) = \frac{2 \tan A}{1 - \tan^2 A \tan^2 B}$ is not standard,so we expand:
$= \left( \frac{1 + \tan \theta}{1 - \tan \theta} \right) + \left( \frac{1 - \tan \theta}{1 + \tan \theta} \right)$
$= \frac{(1 + \tan \theta)^2 + (1 - \tan \theta)^2}{(1 - \tan \theta)(1 + \tan \theta)}$
$= \frac{1 + 2 \tan \theta + \tan^2 \theta + 1 - 2 \tan \theta + \tan^2 \theta}{1 - \tan^2 \theta}$
$= \frac{2(1 + \tan^2 \theta)}{1 - \tan^2 \theta} = \frac{2}{\cos 2 \theta}$.
Substituting $\cos 2 \theta = \frac{a}{b}$,we get $\frac{2}{a/b} = \frac{2b}{a}$.

(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$.
Taking $\sin$ on both sides:
$\sin(\sin ^{-1} x + \sin ^{-1}(1-x)) = \sin(\cos ^{-1} x)$.
Using $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$x \sqrt{1-(1-x)^2} + (1-x) \sqrt{1-x^2} = \sqrt{1-x^2}$.
$x \sqrt{1-(1-2x+x^2)} + (1-x) \sqrt{1-x^2} = \sqrt{1-x^2}$.
$x \sqrt{2x-x^2} + (1-x) \sqrt{1-x^2} = \sqrt{1-x^2}$.
$x \sqrt{x(2-x)} = \sqrt{1-x^2} - (1-x) \sqrt{1-x^2} = x \sqrt{1-x^2}$.
Squaring both sides: $x^2(2x-x^2) = x^2(1-x^2)$.
$2x^3 - x^4 = x^2 - x^4$.
$2x^3 - x^2 = 0 \Rightarrow x^2(2x-1) = 0$.
So,$x = 0$ or $x = 1/2$.
Checking $x=0$: $\sin^{-1}(0) + \sin^{-1}(1) = 0 + \pi/2 = \pi/2$. $\cos^{-1}(0) = \pi/2$. (Valid)
Checking $x=1/2$: $\sin^{-1}(1/2) + \sin^{-1}(1/2) = \pi/6 + \pi/6 = \pi/3$. $\cos^{-1}(1/2) = \pi/3$. (Valid)
Thus,there are $2$ solutions.

(C) Let $S = \tan ^{-1}\left(\frac{1}{2} \tan 2 A\right)+\tan ^{-1}(\cot A)+\tan ^{-1}(\cot ^{3} A)$.
Using the identity $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$,we have:
$\tan ^{-1}(\cot A) + \tan ^{-1}(\cot ^{3} A) = \tan ^{-1}\left(\frac{\cot A + \cot ^{3} A}{1 - \cot A \cdot \cot ^{3} A}\right) = \tan ^{-1}\left(\frac{\cot A(1 + \cot ^{2} A)}{1 - \cot ^{4} A}\right)$.
Since $1 - \cot ^{4} A = (1 - \cot ^{2} A)(1 + \cot ^{2} A)$,the expression simplifies to:
$\tan ^{-1}\left(\frac{\cot A}{1 - \cot ^{2} A}\right) = \tan ^{-1}\left(\frac{1/\tan A}{1 - 1/\tan ^{2} A}\right) = \tan ^{-1}\left(\frac{\tan A}{\tan ^{2} A - 1}\right) = -\tan ^{-1}\left(\frac{\tan A}{1 - \tan ^{2} A}\right)$.
Now,$\frac{1}{2} \tan 2 A = \frac{1}{2} \cdot \frac{2 \tan A}{1 - \tan ^{2} A} = \frac{\tan A}{1 - \tan ^{2} A}$.
Thus,$S = \tan ^{-1}\left(\frac{\tan A}{1 - \tan ^{2} A}\right) - \tan ^{-1}\left(\frac{\tan A}{1 - \tan ^{2} A}\right) = 0$.

(A) Given the equation: $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$\tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}\right) \left(\frac{x+1}{x+2}\right)}\right]=\frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{(x-1)(x+1)}{(x-2)(x+2)}} = \tan \frac{\pi}{4} = 1$
Simplifying the numerator and denominator:
$\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)} = 1$
$\frac{(x^2+x-2)+(x^2-x-2)}{(x^2-4)-(x^2-1)} = 1$
$\frac{2x^2-4}{-3} = 1$
$2x^2-4 = -3$
$2x^2 = 1$
$x^2 = \frac{1}{2}$
$x = \pm \frac{1}{\sqrt{2}}$

(B) Let $\theta = \sin^{-1}\left(\frac{2}{\sqrt{13}}\right)$ and $\phi = \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)$.
Then $\sin \theta = \frac{2}{\sqrt{13}}$,which implies $\tan \theta = \frac{2}{3}$.
Then $\cos \phi = \frac{3}{\sqrt{10}}$,which implies $\tan \phi = \frac{1}{3}$.
We need to find $\tan(2\theta - 2\phi) = \frac{\tan 2\theta - \tan 2\phi}{1 + \tan 2\theta \tan 2\phi}$.
Using $\tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}$:
$\tan 2\theta = \frac{2(2/3)}{1 - (2/3)^2} = \frac{4/3}{1 - 4/9} = \frac{4/3}{5/9} = \frac{4}{3} \times \frac{9}{5} = \frac{12}{5}$.
$\tan 2\phi = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}$.
Substituting these values:
$\tan(2\theta - 2\phi) = \frac{12/5 - 3/4}{1 + (12/5)(3/4)} = \frac{(48-15)/20}{1 + 36/20} = \frac{33/20}{56/20} = \frac{33}{56}$.

(A) Let $\alpha = \frac{1}{2}\cos^{-1}(\frac{2}{3})$ and $\beta = \frac{1}{2}\sin^{-1}(\frac{2}{3})$.
Then $k = \tan(\frac{\pi}{4} + \alpha) + \tan(\beta)$.
Since $2\alpha = \cos^{-1}(\frac{2}{3})$,$\cos(2\alpha) = \frac{2}{3}$.
Since $2\beta = \sin^{-1}(\frac{2}{3})$,$\sin(2\beta) = \frac{2}{3}$.
Note that $\beta = \frac{\pi}{4} - \alpha$ is not necessarily true,but $\tan(\frac{\pi}{4} + \alpha) = \frac{1 + \tan \alpha}{1 - \tan \alpha}$.
Using $\tan \alpha = \sqrt{\frac{1-\cos 2\alpha}{1+\cos 2\alpha}} = \sqrt{\frac{1-2/3}{1+2/3}} = \sqrt{\frac{1/3}{5/3}} = \frac{1}{\sqrt{5}}$.
Using $\tan \beta = \sqrt{\frac{1-\cos 2\beta}{1+\cos 2\beta}} = \sqrt{\frac{1-\sqrt{1-4/9}}{1+\sqrt{1-4/9}}} = \sqrt{\frac{1-\sqrt{5}/3}{1+\sqrt{5}/3}} = \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}} = \frac{3-\sqrt{5}}{2}$.
Calculating $k$ leads to $k=3$.
The equation becomes $\sin^{-1}(3x-1) = \sin^{-1}x - (\frac{\pi}{2} - \sin^{-1}x) = 2\sin^{-1}x - \frac{\pi}{2}$.
Taking $\sin$ on both sides: $3x-1 = \sin(2\sin^{-1}x - \frac{\pi}{2}) = -\cos(2\sin^{-1}x) = -(1-2x^2) = 2x^2-1$.
$2x^2 - 3x = 0 \Rightarrow x(2x-3) = 0$.
So $x=0$ or $x=1.5$.
Checking $x=1.5$: $\sin^{-1}(3.5)$ is undefined.
Checking $x=0$: $\sin^{-1}(-1) = -\frac{\pi}{2}$ and $\sin^{-1}(0) - \cos^{-1}(0) = 0 - \frac{\pi}{2} = -\frac{\pi}{2}$.
Thus,$x=0$ is the only solution.

(C) Let $u = \sqrt{x^2+x}$. The domain requires $x^2+x \ge 0$ and $0 \le x^2+x+1 \le 1$.
Since $x^2+x+1 \le 1 \Rightarrow x^2+x \le 0$.
Combining $x^2+x \ge 0$ and $x^2+x \le 0$,we must have $x^2+x = 0$.
If $x^2+x = 0$,then $u = 0$.
The equation becomes $\tan^{-1}(0) + \sin^{-1}(1) = 0 + \pi/2 = \pi/2$.
This satisfies the equation.
$x^2+x = 0 \Rightarrow x(x+1) = 0$,which gives $x = 0$ or $x = -1$.
Both values are valid.
Thus,there are $2$ real solutions.

(A) Let $\tan^{-1}(x\sqrt{2}) = \theta$. Then $\tan \theta = x\sqrt{2}$.
Using the identity $\sin \theta = \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}}$,we get $\sin \theta = \frac{x\sqrt{2}}{\sqrt{1+2x^2}}$.
Let $\sin^{-1}(\sqrt{1-x^2}) = \phi$. Then $\sin \phi = \sqrt{1-x^2}$.
Using the identity $\cot \phi = \frac{\cos \phi}{\sin \phi} = \frac{\sqrt{1-\sin^2 \phi}}{\sin \phi}$,we get $\cot \phi = \frac{\sqrt{1-(1-x^2)}}{\sqrt{1-x^2}} = \frac{x}{\sqrt{1-x^2}}$.
Equating the two sides: $\frac{x\sqrt{2}}{\sqrt{1+2x^2}} = \frac{x}{\sqrt{1-x^2}}$.
Since $x \in (0, 1)$,$x \neq 0$,we can divide by $x$: $\frac{\sqrt{2}}{\sqrt{1+2x^2}} = \frac{1}{\sqrt{1-x^2}}$.
Squaring both sides: $\frac{2}{1+2x^2} = \frac{1}{1-x^2}$.
$2(1-x^2) = 1+2x^2 \implies 2-2x^2 = 1+2x^2 \implies 4x^2 = 1 \implies x^2 = 1/4$.
Since $x \in (0, 1)$,$x = 1/2$.

(A) Given $\alpha = 3 \sin^{-1}(\frac{6}{11})$. Since $\frac{6}{11} > \frac{1}{2}$,we have $\sin^{-1}(\frac{6}{11}) > \sin^{-1}(\frac{1}{2}) = 30^\circ$. Thus,$\alpha > 3 \times 30^\circ = 90^\circ$. Since $90^\circ < \alpha < 270^\circ$ (as $\sin^{-1}(\frac{6}{11}) < 90^\circ$),$\cos(\alpha) < 0$. So,Statement $II$ is true.
Given $\beta = 3 \cos^{-1}(\frac{4}{9})$. Since $\frac{4}{9} < \frac{1}{2}$,we have $\cos^{-1}(\frac{4}{9}) > \cos^{-1}(\frac{1}{2}) = 60^\circ$. Thus,$\beta > 3 \times 60^\circ = 180^\circ$.
Since $\alpha > 90^\circ$ and $\beta > 180^\circ$,$\alpha + \beta > 270^\circ$. In the fourth quadrant,the cosine function is positive. Therefore,$\cos(\alpha + \beta) > 0$. So,Statement $I$ is true.