Basic Concepts of ITF Questions in English

(A) Given the partial fraction decomposition: $\frac{(x + 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.
Multiplying both sides by $x(x^2 + 1)$,we get: $(x + 1)^2 = A(x^2 + 1) + (Bx + C)x$.
Expanding the left side: $x^2 + 2x + 1 = Ax^2 + A + Bx^2 + Cx$.
Grouping terms: $x^2 + 2x + 1 = (A + B)x^2 + Cx + A$.
Comparing coefficients:
For $x^2$: $A + B = 1$.
For $x$: $C = 2$.
Constant term: $A = 1$.
Substituting $A = 1$ into $A + B = 1$,we get $1 + B = 1$,so $B = 0$.
Thus,$A = 1$ and $C = 2$.
We need to find $\sin^{-1}\left(\frac{A}{C}\right) = \sin^{-1}\left(\frac{1}{2}\right)$.
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,the value is $\frac{\pi}{6}$.

(A) Let $\sin^{-1}(e^{i\theta}) = x + iy$.
Then,$e^{i\theta} = \sin(x + iy)$.
Using the identity $\sin(x + iy) = \sin x \cosh y + i \cos x \sinh y$,we have:
$\sin x \cosh y + i \cos x \sinh y = \cos \theta + i \sin \theta$.
Comparing real and imaginary parts:
$\sin x \cosh y = \cos \theta$ and $\cos x \sinh y = \sin \theta$.
Thus,$\cosh y = \frac{\cos \theta}{\sin x}$ and $\sinh y = \frac{\sin \theta}{\cos x}$.
Using the identity $\cosh^2 y - \sinh^2 y = 1$:
$\frac{\cos^2 \theta}{\sin^2 x} - \frac{\sin^2 \theta}{\cos^2 x} = 1$.
$\cos^2 \theta \cos^2 x - \sin^2 \theta \sin^2 x = \sin^2 x \cos^2 x$.
$\cos^2 \theta \cos^2 x - \sin^2 \theta (1 - \cos^2 x) = (1 - \cos^2 x) \cos^2 x$.
$\cos^2 \theta \cos^2 x - \sin^2 \theta + \sin^2 \theta \cos^2 x = \cos^2 x - \cos^4 x$.
$\cos^2 x (\cos^2 \theta + \sin^2 \theta) - \sin^2 \theta = \cos^2 x - \cos^4 x$.
$\cos^2 x - \sin^2 \theta = \cos^2 x - \cos^4 x$.
$\cos^4 x = \sin^2 \theta$.
Taking the square root,$\cos^2 x = \sin \theta$,so $x = \cos^{-1}(\sqrt{\sin \theta})$.

(B) Given that $\tan \theta = -\frac{1}{\sqrt{3}}$,$\sin \theta = \frac{1}{2}$,and $\cos \theta = -\frac{\sqrt{3}}{2}$.
Since $\sin \theta > 0$ and $\cos \theta < 0$,the angle $\theta$ must lie in the second quadrant.
The reference angle $\alpha$ is given by $\tan \alpha = \frac{1}{\sqrt{3}}$,which implies $\alpha = \frac{\pi}{6}$.
In the second quadrant,$\theta = \pi - \alpha = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Thus,the principal value of $\theta$ is $\frac{5\pi}{6}$.

(D) Given that ${\cos ^{ - 1}}\left( {\frac{1}{x}} \right) = \theta $.
This implies $\cos \theta = \frac{1}{x}$.
We know that $\sin^2 \theta + \cos^2 \theta = 1$,so $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{1}{x}\right)^2} = \sqrt{\frac{x^2 - 1}{x^2}} = \frac{\sqrt{x^2 - 1}}{x}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{x^2 - 1}}{x}}{\frac{1}{x}} = \sqrt{x^2 - 1}$.
Therefore,the correct option is $D$.

(D) Let $\theta = \cot ^{ - 1}x$. Then,$\cot \theta = x = \frac{x}{1}$.
We know that $\csc^2 \theta = 1 + \cot^2 \theta = 1 + x^2$.
Therefore,$\csc \theta = \sqrt{1 + x^2}$.
Since $\sin \theta = \frac{1}{\csc \theta}$,we have $\sin \theta = \frac{1}{\sqrt{1 + x^2}}$.
Thus,$\sin (\cot ^{ - 1}x) = \frac{1}{\sqrt{1 + x^2}} = (1 + x^2)^{-1/2}$.

(A) Let $\sin^{-1} \frac{5}{13} = x$.
Then,$\sin x = \frac{5}{13}$.
Since $\cos^2 x = 1 - \sin^2 x$,we have $\cos x = \sqrt{1 - \left( \frac{5}{13} \right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
Therefore,$\cos \left( \sin^{-1} \frac{5}{13} \right) = \cos x = \frac{12}{13}$.
Since the value $\frac{12}{13}$ is present in option $A$,the correct answer is $A$.

(B) We know that for the inverse cotangent function,the range is $(0, \pi)$.
Since $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for $x > 0$,we have:
$\cot^{-1}(-\sqrt{3}) = \pi - \cot^{-1}(\sqrt{3})$
We know that $\cot(\frac{\pi}{6}) = \sqrt{3}$,so $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$.
Therefore,$\cot^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

(C) Let $\sin^{-1}x = \theta$.
Then,$\sin \theta = x$.
We know that $1 + \cot^2 \theta = \csc^2 \theta$.
Since $\csc \theta = \frac{1}{\sin \theta}$,we have $\csc^2 \theta = \frac{1}{\sin^2 \theta}$.
Substituting $\sin \theta = x$,we get $\csc^2 \theta = \frac{1}{x^2}$.
Therefore,$1 + \cot^2(\sin^{-1}x) = \frac{1}{x^2}$.

(B) Given the equation: ${\sin ^{ - 1}}\frac{1}{2} = {\tan ^{ - 1}}x$
We know that ${\sin ^{ - 1}}\frac{1}{2} = \frac{\pi}{6}$ because $\sin \frac{\pi}{6} = \frac{1}{2}$.
Substituting this into the equation,we get: $\frac{\pi}{6} = {\tan ^{ - 1}}x$
Taking the tangent of both sides: $x = \tan \frac{\pi}{6}$
Since $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$,we have $x = \frac{1}{\sqrt{3}}$.

(C) To evaluate $\tan ^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right)$,we use the substitution $x = a \sin \theta$.
Then,$\sin \theta = \frac{x}{a}$,which implies $\theta = \sin ^{-1} \left( \frac{x}{a} \right)$.
Substituting $x = a \sin \theta$ into the expression:
$\tan ^{-1} \left( \frac{a \sin \theta}{\sqrt{a^2 - (a \sin \theta)^2}} \right) = \tan ^{-1} \left( \frac{a \sin \theta}{\sqrt{a^2(1 - \sin^2 \theta)}} \right)$
$= \tan ^{-1} \left( \frac{a \sin \theta}{\sqrt{a^2 \cos^2 \theta}} \right) = \tan ^{-1} \left( \frac{a \sin \theta}{a \cos \theta} \right)$
$= \tan ^{-1} (\tan \theta) = \theta$
Substituting back the value of $\theta$,we get $\sin ^{-1} \left( \frac{x}{a} \right)$.

(B) Let $\theta = \tan ^{ - 1}x$.
Then,$x = \tan \theta$.
We know that $\cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1 + \tan^2 \theta}}$.
Substituting $x = \tan \theta$ into the expression,we get:
$\cos \theta = \frac{1}{\sqrt{1 + x^2}}$.
Therefore,$\cos (\tan ^{ - 1}x) = \frac{1}{\sqrt{1 + x^2}}$.

(B) Let $x = \tan \theta$,where $\theta = \tan^{-1} x$.
Then,$\sqrt{1 + x^2} = \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta$.
Substituting this into the expression:
$\tan \left[ \sec^{-1} \sqrt{1 + x^2} \right] = \tan \left[ \sec^{-1} (\sec \theta) \right]$.
Since $\sec^{-1} (\sec \theta) = \theta$,we have:
$\tan \theta = x$.
Therefore,the correct option is $B$.

(C) We know that $\sec(-x) = \sec(x)$.
Therefore,$\sec(-30^o) = \sec(30^o)$.
Now,the expression becomes $\sec^{-1}(\sec(30^o))$.
Since the principal value branch of $\sec^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$,and $30^o$ lies within this range,
$\sec^{-1}(\sec(30^o)) = 30^o$.

(C) Let $\tan^{-1} 2 = \alpha \implies \tan \alpha = 2$.
Let $\cot^{-1} 3 = \beta \implies \cot \beta = 3$.
The given expression is $\sec^2 \alpha + \csc^2 \beta$.
Using the trigonometric identities $\sec^2 \theta = 1 + \tan^2 \theta$ and $\csc^2 \theta = 1 + \cot^2 \theta$,we get:
$\sec^2 \alpha + \csc^2 \beta = (1 + \tan^2 \alpha) + (1 + \cot^2 \beta)$.
Substituting the values $\tan \alpha = 2$ and $\cot \beta = 3$:
$= (1 + 2^2) + (1 + 3^2)$
$= (1 + 4) + (1 + 9)$
$= 5 + 10 = 15$.

(B) The range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.
Since $\frac{7\pi}{6} > \pi$,we cannot directly write $\cos^{-1}(\cos \theta) = \theta$.
We use the property $\cos(\pi + \theta) = -\cos \theta$ and $\cos^{-1}(-x) = \pi - \cos^{-1} x$.
$\cos^{-1} \left( \cos \frac{7\pi}{6} \right) = \cos^{-1} \left( \cos \left( \pi + \frac{\pi}{6} \right) \right)$
$= \cos^{-1} \left( -\cos \frac{\pi}{6} \right)$
$= \pi - \cos^{-1} \left( \cos \frac{\pi}{6} \right)$
$= \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

(A) Let $\cos ^{ - 1}x = \theta$. Then $x = \cos \theta$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta}$.
Substituting the values,we get $\tan \theta = \frac{\sqrt{1 - x^2}}{x}$.
Therefore,$\tan (\cos ^{ - 1}x) = \frac{\sqrt{1 - x^2}}{x}$.

(B) We know that the range of the principal value branch of ${\sin ^{ - 1}}x$ is $\left[ { - \frac{\pi }{2}, \frac{\pi }{2}} \right]$.
Since ${\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x$ for all $x \in [ - 1, 1]$,we have:
${\sin ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right) = - {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right)$.
We know that $\sin \left( \frac{\pi }{3} \right) = \frac{{\sqrt 3 }}{2}$,therefore ${\sin ^{ - 1}}\left( \frac{{\sqrt 3 }}{2} \right) = \frac{\pi }{3}$.
Thus,${\sin ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right) = - \frac{\pi }{3}$.

(D) Let $\theta = \cos^{-1} \left( \frac{7}{25} \right)$.
Then $\cos \theta = \frac{7}{25}$.
Since $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{25}$,we can find the opposite side using the Pythagorean theorem: $\text{opposite} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24$.
Therefore,$\cot \theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{7}{24}$.
Thus,$\cot \left[ \cos^{-1} \left( \frac{7}{25} \right) \right] = \frac{7}{24}$.

(D) We know that the range of the principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Given $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$.
We use the property $\sin x = \sin(\pi - x)$.
Since $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$,we have $-\frac{\pi}{2} \le \pi - x \le \frac{\pi}{2}$.
Therefore,$\sin^{-1}(\sin x) = \sin^{-1}(\sin(\pi - x)) = \pi - x$.

(C) We know that $\sin^{-1}(\sin x) = x$ only when $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $10$ radians is not in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we need to find an angle $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(y) = \sin(10)$.
We know that $\sin(x) = \sin(n\pi + (-1)^n x)$.
For $n = 3$,$\sin(10) = \sin(3\pi - 10)$.
Since $3\pi \approx 9.42$,we have $3\pi < 10 < 3.5\pi$.
Specifically,$3\pi < 10 < 3\pi + \frac{\pi}{2}$.
Subtracting $3\pi$ from all sides,we get $0 < 10 - 3\pi < \frac{\pi}{2}$.
Multiplying by $-1$,we get $-\frac{\pi}{2} < 3\pi - 10 < 0$.
Since $3\pi - 10$ lies in the principal value branch $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\sin^{-1}(\sin 10) = \sin^{-1}(\sin(3\pi - 10)) = 3\pi - 10$.

(C) Let $\theta = \tan^{-1}\left(\frac{3}{4}\right)$.
Then,$\tan \theta = \frac{3}{4}$.
In a right-angled triangle,if the opposite side is $3$ and the adjacent side is $4$,then the hypotenuse is $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
Therefore,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$.
Substituting this back into the expression: ${\left[ \sin \left( \tan^{-1} \frac{3}{4} \right) \right]^2} = (\sin \theta)^2 = \left( \frac{3}{5} \right)^2 = \frac{9}{25}$.

(C) We know that the range of the principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Given expression is $\sin^{-1} \left( \sin \frac{5\pi}{3} \right)$.
First,simplify the argument: $\sin \frac{5\pi}{3} = \sin \left( 2\pi - \frac{\pi}{3} \right) = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}$.
Now,$\sin^{-1} \left( -\frac{\sqrt{3}}{2} \right) = -\sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = -\frac{\pi}{3}$.
Since $-\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the principal value is $-\frac{\pi}{3}$.

(A) Given the equation: $\tan^{-1}x = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right)$.
Let $\theta = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right)$. Then $\sin\theta = \frac{3}{\sqrt{10}}$.
Using the identity $\cos^2\theta = 1 - \sin^2\theta$,we get $\cos^2\theta = 1 - \left(\frac{3}{\sqrt{10}}\right)^2 = 1 - \frac{9}{10} = \frac{1}{10}$.
Thus,$\cos\theta = \frac{1}{\sqrt{10}}$.
Now,$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/\sqrt{10}}{1/\sqrt{10}} = 3$.
Therefore,$\theta = \tan^{-1}(3)$.
Substituting this back into the original equation: $\tan^{-1}x = \tan^{-1}(3)$.
Hence,$x = 3$.

(A) Given $\theta = \sin^{-1}[\sin(-600^\circ)]$.
Since $\sin(-\theta) = -\sin(\theta)$,we have $\theta = \sin^{-1}[-\sin(600^\circ)]$.
Now,express $600^\circ$ as a multiple of $180^\circ$ or $360^\circ$: $600^\circ = 360^\circ + 240^\circ$.
So,$\sin(600^\circ) = \sin(360^\circ + 240^\circ) = \sin(240^\circ)$.
Also,$\sin(240^\circ) = \sin(180^\circ + 60^\circ) = -\sin(60^\circ)$.
Substituting this back: $\theta = \sin^{-1}[-(-\sin(60^\circ))] = \sin^{-1}[sin(60^\circ)]$.
Since $60^\circ$ is in the principal value branch $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\theta = 60^\circ = \frac{\pi}{3}$.

(A) Given the equation: $\sin (\cot ^{ - 1}(x + 1)) = \cos (\tan ^{ - 1}x)$.
First,convert $\cot ^{ - 1}(x + 1)$ to $\sin ^{ - 1}$ form. Let $\theta = \cot ^{ - 1}(x + 1)$,then $\cot \theta = x + 1$. Using the identity $\sin \theta = \frac{1}{\sqrt{1 + \cot ^2 \theta}}$,we get $\sin \theta = \frac{1}{\sqrt{1 + (x + 1)^2}} = \frac{1}{\sqrt{x^2 + 2x + 2}}$.
Next,convert $\tan ^{ - 1}x$ to $\cos ^{ - 1}$ form. Let $\phi = \tan ^{ - 1}x$,then $\tan \phi = x$. Using the identity $\cos \phi = \frac{1}{\sqrt{1 + \tan ^2 \phi}}$,we get $\cos \phi = \frac{1}{\sqrt{1 + x^2}}$.
Equating the two sides: $\frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{1 + x^2}}$.
Squaring both sides: $x^2 + 2x + 2 = 1 + x^2$.
Subtracting $x^2$ from both sides: $2x + 2 = 1$.
Solving for $x$: $2x = -1$,which gives $x = -\frac{1}{2}$.

(C) We know that $\tan(90^o - \theta) = \cot(\theta)$.
Substituting $\theta = \cot^{-1} \frac{1}{3}$,we get:
$\tan \left( 90^o - \cot^{-1} \frac{1}{3} \right) = \cot \left( \cot^{-1} \frac{1}{3} \right)$.
Since $\cot(\cot^{-1} x) = x$ for all $x \in \mathbb{R}$,we have:
$\cot \left( \cot^{-1} \frac{1}{3} \right) = \frac{1}{3}$.
Thus,the correct option is $C$.

(B) We know the identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$ for $\theta \in [-1, 1]$.
In the given expression,let $\theta = \sqrt{1-x}$.
For the expression to be defined,we must have $0 \le 1-x \le 1$,which implies $0 \le x \le 1$.
Thus,$\cos^{-1}\sqrt{1-x} + \sin^{-1}\sqrt{1-x} = \frac{\pi}{2}$.

(C) We know the identity ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$.
Given the equation ${\cot ^{ - 1}}x + {\tan ^{ - 1}}3 = \frac{\pi }{2}$.
Comparing this with the identity ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$,we can see that the values must match.
Therefore,$x = 3$.

(D) We know that $\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$ because $\cos(\frac{\pi}{3}) = \frac{1}{2}$.
We also know that $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ because $\sin(\frac{\pi}{6}) = \frac{1}{2}$.
Substituting these values into the expression:
$\cos^{-1}(\frac{1}{2}) + 2\sin^{-1}(\frac{1}{2}) = \frac{\pi}{3} + 2(\frac{\pi}{6})$
$= \frac{\pi}{3} + \frac{\pi}{3}$
$= \frac{2\pi}{3}$.

(A) We know that $\cos^{-1}(\cos x) = x$ for $x \in [0, \pi]$ and $\sin^{-1}(\sin x) = x$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
First,simplify $\cos^{-1}(\cos \frac{5\pi}{3})$:
$\cos \frac{5\pi}{3} = \cos(2\pi - \frac{\pi}{3}) = \cos(\frac{\pi}{3})$.
Since $\frac{\pi}{3} \in [0, \pi]$,$\cos^{-1}(\cos \frac{5\pi}{3}) = \frac{\pi}{3}$.
Next,simplify $\sin^{-1}(\sin \frac{5\pi}{3})$:
$\sin \frac{5\pi}{3} = \sin(2\pi - \frac{\pi}{3}) = \sin(-\frac{\pi}{3})$.
Since $-\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$\sin^{-1}(\sin \frac{5\pi}{3}) = -\frac{\pi}{3}$.
Adding these values:
$\frac{\pi}{3} + (-\frac{\pi}{3}) = 0$.

(D) We know that $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^o$ because $\sin(60^o) = \frac{\sqrt{3}}{2}$.
We also know that $\sin^{-1}\left(\frac{1}{2}\right) = 30^o$ because $\sin(30^o) = \frac{1}{2}$.
Therefore,$\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) - \sin^{-1}\left(\frac{1}{2}\right) = 60^o - 30^o = 30^o$.

(C) Given that $A = \tan^{-1}x$.
From this,we can write $x = \tan A$.
We know the trigonometric identity for $\sin 2A$ in terms of $\tan A$ is:
$\sin 2A = \frac{2\tan A}{1 + \tan^2 A}$.
Substituting $x = \tan A$ into the identity,we get:
$\sin 2A = \frac{2x}{1 + x^2}$.
Therefore,the correct option is $C$.

(C) Let $\theta = \cos^{-1}\left(\frac{3}{5}\right)$. Then $\cos \theta = \frac{3}{5}$.
Since $\cos \theta = \frac{3}{5}$,we have $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Thus,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{4/5}{3/5} = \frac{4}{3}$.
We need to find $\tan(2\theta)$.
Using the formula $\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$,we get:
$\tan(2\theta) = \frac{2(4/3)}{1 - (4/3)^2} = \frac{8/3}{1 - 16/9} = \frac{8/3}{-7/9} = \frac{8}{3} \times \left(-\frac{9}{7}\right) = -\frac{24}{7}$.

(B) Given equation is $2\cos^{-1}\sqrt{\frac{1+x}{2}} = \frac{\pi}{2}.$
Dividing both sides by $2,$ we get $\cos^{-1}\sqrt{\frac{1+x}{2}} = \frac{\pi}{4}.$
Taking cosine on both sides,we have $\sqrt{\frac{1+x}{2}} = \cos\left(\frac{\pi}{4}\right).$
Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}},$ we get $\sqrt{\frac{1+x}{2}} = \frac{1}{\sqrt{2}}.$
Squaring both sides,$\frac{1+x}{2} = \frac{1}{2}.$
Multiplying by $2,$ we get $1+x = 1,$
which implies $x = 0.$

(A) Let $\theta = \frac{1}{2} \cos^{-1} \left( \frac{\sqrt{5}}{3} \right)$.
Then $2\theta = \cos^{-1} \left( \frac{\sqrt{5}}{3} \right)$,which implies $\cos 2\theta = \frac{\sqrt{5}}{3}$.
Using the formula $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$,we have:
$\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{\sqrt{5}}{3}$.
Cross-multiplying gives $3 - 3 \tan^2 \theta = \sqrt{5} + \sqrt{5} \tan^2 \theta$.
Rearranging terms: $(3 + \sqrt{5}) \tan^2 \theta = 3 - \sqrt{5}$.
$\tan^2 \theta = \frac{3 - \sqrt{5}}{3 + \sqrt{5}}$.
Rationalizing the denominator: $\tan^2 \theta = \frac{(3 - \sqrt{5})^2}{9 - 5} = \frac{(3 - \sqrt{5})^2}{4}$.
Taking the square root: $\tan \theta = \frac{3 - \sqrt{5}}{2}$.

(A) Let $\cos^{-1} \frac{4}{5} = x$,which implies $\cos x = \frac{4}{5}$.
We need to find $\sin \left( \frac{x}{2} \right)$.
Using the trigonometric identity $\cos x = 1 - 2 \sin^2 \left( \frac{x}{2} \right)$:
$\frac{4}{5} = 1 - 2 \sin^2 \left( \frac{x}{2} \right)$
$2 \sin^2 \left( \frac{x}{2} \right) = 1 - \frac{4}{5} = \frac{1}{5}$
$\sin^2 \left( \frac{x}{2} \right) = \frac{1}{10}$
Since $\cos^{-1} \frac{4}{5}$ is in the first quadrant,$\frac{x}{2}$ is also in the first quadrant,so $\sin \left( \frac{x}{2} \right)$ must be positive.
Therefore,$\sin \left( \frac{x}{2} \right) = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$.

(D) The principal value branch of ${\sin^{-1}}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\frac{2\pi}{3}$ does not lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we simplify the expression.
We know that $\sin(\pi - \theta) = \sin(\theta)$.
Therefore,$\sin(\frac{2\pi}{3}) = \sin(\pi - \frac{\pi}{3}) = \sin(\frac{\pi}{3})$.
Now,${\sin^{-1}}[\sin(\frac{2\pi}{3})] = {\sin^{-1}}[\sin(\frac{\pi}{3})]$.
Since $\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the value is $\frac{\pi}{3}$.

(C) We know that the identity ${\sin ^{ - 1}} \theta + {\cos ^{ - 1}} \theta = \frac{\pi }{2}$ is valid for all $\theta \in [-1, 1]$.
In the given expression,$\theta = \sqrt x$.
Therefore,the condition becomes $-1 \le \sqrt x \le 1$.
Since $\sqrt x$ is always non-negative,we have $0 \le \sqrt x \le 1$.
Squaring all sides,we get $0^2 \le x \le 1^2$,which simplifies to $0 \le x \le 1$.
Thus,the interval for which the equation holds is $x \in [0, 1]$.

(B) Given $f(x) = 1$ for $x > 0$. Since $x = \frac{5\pi}{4} > 0$,we have $f(x) = 1$.
Thus,$y = \cos^{-1}(\cos(|x| - 1))$.
Since $x = \frac{5\pi}{4} > 0$,$|x| = x$,so $y = \cos^{-1}(\cos(x - 1))$.
For $x = \frac{5\pi}{4}$,$x - 1 = \frac{5\pi}{4} - 1 \approx 3.927 - 1 = 2.927$.
Since $0 \le 2.927 \le \pi$ (where $\pi \approx 3.14159$),the expression simplifies to $y = x - 1$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x - 1) = 1$.
Evaluating at $x = \frac{5\pi}{4}$,we get $\left. \frac{dy}{dx} \right|_{x = \frac{5\pi}{4}} = 1$.

(A) Given $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$.
Substitute $x = \tan \theta$,which implies $\theta = \tan^{-1} x$.
Since $0 < x < 1$,we have $0 < \tan \theta < 1$,which implies $0 < \theta < \frac{\pi}{4}$,and thus $0 < 2\theta < \frac{\pi}{2}$.
Using the trigonometric identity $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we get:
$y = \sin^{-1} (\sin 2\theta) = 2\theta$.
Substituting back $\theta = \tan^{-1} x$,we have $y = 2 \tan^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2}$.

(B) Given the equation $x + \frac{1}{x} = 2$.
Multiplying by $x$,we get $x^2 + 1 = 2x$,which simplifies to $x^2 - 2x + 1 = 0$.
This is a perfect square: $(x - 1)^2 = 0$,which implies $x = 1$.
We need to find the principal value of $\sin^{-1} x$.
Substituting $x = 1$,we get $\sin^{-1}(1)$.
Since $\sin(\pi / 2) = 1$,the principal value is $\pi / 2$.

(A) Let $\theta = \sin^{-1}(0.8)$. Then $\sin \theta = 0.8 = \frac{4}{5}$.
We need to find $\sin(2\theta)$.
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$.
Since $\sin \theta = \frac{4}{5}$,we have $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Therefore,$\sin(2\theta) = 2 \times \frac{4}{5} \times \frac{3}{5} = \frac{24}{25} = 0.96$.

(B) The function $f(x) = \sin^{-1}(\sin x)$ is a periodic function with period $2\pi$.
It is defined as:
$f(x) = \begin{cases} x - 2k\pi, & 2k\pi - \frac{\pi}{2} \le x \le 2k\pi + \frac{\pi}{2} \\ (2k+1)\pi - x, & 2k\pi + \frac{\pi}{2} < x < 2k\pi + \frac{3\pi}{2} \end{cases}$
where $k \in I$.
Since $\sin^{-1}(\sin x)$ is a continuous function for all $x \in R$,it is continuous everywhere.
However,the function is not differentiable at points where the slope changes abruptly,which are the points $x = (2k + 1)\frac{\pi}{2}$ for all $k \in I$.
Thus,the function is continuous for all $x$ but not differentiable at $x = (2k + 1)\frac{\pi}{2}$.

(C) We know that $\mathop {Lim}\limits_{x \to 0} \frac{\sin x}{x} = 1$,so $\mathop {Lim}\limits_{x \to 0} \frac{x}{\sin x} = 1$.
Thus,$l = \sec^{-1}(1) = 0$.
Similarly,$\mathop {Lim}\limits_{x \to 0} \frac{\tan x}{x} = 1$,so $\mathop {Lim}\limits_{x \to 0} \frac{x}{\tan x} = 1$.
Thus,$m = \sec^{-1}(1) = 0$.
Since both limits exist and are equal to $0$,$l$ and $m$ both exist.

(B) To find the number of solutions for $|\sin^{-1}x| = |x|$,we analyze the graphs of $f(x) = |\sin^{-1}x|$ and $g(x) = |x|$.
$1$. The domain of $\sin^{-1}x$ is $[-1, 1]$.
$2$. At $x = 0$,both $|\sin^{-1}0| = 0$ and $|0| = 0$. Thus,$x = 0$ is a solution.
$3$. For $x \in (0, 1]$,we know that $\sin^{-1}x > x$. Since both sides are positive,$|\sin^{-1}x| = \sin^{-1}x$ and $|x| = x$. Thus,$\sin^{-1}x = x$ has no solution for $x > 0$ because $\sin^{-1}x$ is strictly greater than $x$ for all $x \in (0, 1]$.
$4$. For $x \in [-1, 0)$,we have $|\sin^{-1}x| = |-\sin^{-1}(-x)| = |\sin^{-1}(-x)|$. Since $-x > 0$,$\sin^{-1}(-x) > -x$,so $|\sin^{-1}x| > |x|$.
$5$. Therefore,the only point of intersection is at the origin $(0, 0)$.
$6$. The number of solutions is $1$.

(D) To determine the correct order,we evaluate the values of the inverse secant function:
$1$. $\sec^{-1}(1) = 0$
$2$. $\sec^{-1}(2) = \frac{\pi}{3} \approx 1.047$
$3$. $\sec^{-1}(-2) = \frac{2\pi}{3} \approx 2.094$
$4$. $\sec^{-1}(-1) = \pi \approx 3.141$
Comparing these values,we get $0 < 1.047 < 2.094 < 3.141$.
Thus,the correct order is $\sec^{-1}(1) < \sec^{-1}(2) < \sec^{-1}(-2) < \sec^{-1}(-1)$.

(B) Given $f(x) = \sum_{k=0}^{n} \alpha_{2k+1} (\sin^{-1} x)^{2k+1} - \cot^{-1} x$.
Taking the derivative with respect to $x$:
$f'(x) = \frac{1}{\sqrt{1-x^2}} \left( \alpha_1 + 3\alpha_3 (\sin^{-1} x)^2 + \dots + (2n+1)\alpha_{2n+1} (\sin^{-1} x)^{2n} \right) + \frac{1}{1+x^2}$.
Since $\alpha_i > 0$ and the terms $(\sin^{-1} x)^{2k}$ are non-negative,$f'(x) > 0$ for all $x \in (-1, 1)$.
Thus,$f(x)$ is a strictly increasing function,which implies it is one-one.
The domain is $[-1, 1]$,so the range is $[f(-1), f(1)]$.
$f(-1) = \alpha_1(-\pi/2) + \dots + \alpha_{2n+1}(-\pi/2)^{2n+1} - (3\pi/4)$ and $f(1) = \alpha_1(\pi/2) + \dots + \alpha_{2n+1}(\pi/2)^{2n+1} - (\pi/4)$.
Since the range $[f(-1), f(1)]$ is a subset of $R$ but not equal to $R$,the function is into.
Therefore,$f(x)$ is one-one and into.

(B) Let $t = x^2$. Since $x \in R - \{0\}$,$t > 0$.
For $\cos^{-1}(t + \frac{1}{t} - 1)$ to be defined,we must have $-1 \leq t + \frac{1}{t} - 1 \leq 1$.
Since $t + \frac{1}{t} \geq 2$ for $t > 0$,we have $t + \frac{1}{t} - 1 \geq 1$.
Thus,the only possible value is $t + \frac{1}{t} - 1 = 1$,which implies $t + \frac{1}{t} = 2$.
This occurs only when $t = 1$,i.e.,$x^2 = 1$.
Substituting $x^2 = 1$ into the expression:
$\cos^{-1}(1 + 1 - 1) + \sin^{-1}(1 - 1) + \tan^{-1}(1)$
$= \cos^{-1}(1) + \sin^{-1}(0) + \tan^{-1}(1)$
$= 0 + 0 + \frac{\pi}{4} = \frac{\pi}{4}$.

(C) Given equation: $\sin^{-1} 2x = \cos^{-1} x$
Let $\cos^{-1} x = \theta$,then $x = \cos \theta$. Since the range of $\cos^{-1} x$ is $[0, \pi]$,$x$ must be in $[-1, 1]$.
Also,for $\sin^{-1} 2x$ to be defined,$2x \in [-1, 1]$,so $x \in [-\frac{1}{2}, \frac{1}{2}]$.
We know $\cos^{-1} x = \sin^{-1} \sqrt{1-x^2}$ for $x \ge 0$.
So,$\sin^{-1} 2x = \sin^{-1} \sqrt{1-x^2}$
$\Rightarrow 2x = \sqrt{1-x^2}$
Squaring both sides: $4x^2 = 1 - x^2$
$\Rightarrow 5x^2 = 1$
$\Rightarrow x^2 = \frac{1}{5}$
$\Rightarrow x = \pm \frac{1}{\sqrt{5}}$
Check the solutions:
If $x = -\frac{1}{\sqrt{5}}$,then $\sin^{-1}(2(-\frac{1}{\sqrt{5}})) = \sin^{-1}(-\frac{2}{\sqrt{5}})$,which is negative. However,$\cos^{-1}(-\frac{1}{\sqrt{5}})$ is in the second quadrant (positive). Thus,$x = -\frac{1}{\sqrt{5}}$ is not a solution.
If $x = \frac{1}{\sqrt{5}}$,then $\sin^{-1}(\frac{2}{\sqrt{5}}) = \cos^{-1}(\frac{1}{\sqrt{5}})$,which is true.
Therefore,the only solution is $x = \frac{1}{\sqrt{5}}$.

(B) Let $y = \cot^{-1}x$. The inequality becomes $y^2 - 5y + 6 > 0$.
Factoring the quadratic expression,we get $(y - 3)(y - 2) > 0$.
This inequality holds when $y < 2$ or $y > 3$.
Substituting $y = \cot^{-1}x$,we have $\cot^{-1}x < 2$ or $\cot^{-1}x > 3$.
Since the function $\cot^{-1}x$ is a strictly decreasing function,the inequality sign reverses when we apply the cotangent function to both sides.
For $\cot^{-1}x < 2$,we get $x > \cot 2$.
For $\cot^{-1}x > 3$,we get $x < \cot 3$.
Combining these,the solution set is $x \in (-\infty, \cot 3) \cup (\cot 2, \infty)$.