Inverse Trigonometric Functions

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.

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

$\begin{aligned} & \text { (P) Given } y(x)=\cos \left(3 \cos ^{-1} x\right), x \in[-1,1] \\ & \Rightarrow \quad y(x)=4 \cos ^3\left(\cos ^{-1} x\right)-3 \cos \left(\cos ^{-1} x\right) \\ & \Rightarrow \quad y(x)=4 x^3-3 x \quad \text{... (i)}\\ & \Rightarrow \quad \frac{d y(x)}{d x}=12 x^2-3 \quad \text{... (ii)}\\ & \Rightarrow \quad \frac{d^2 y(x)}{d x^2}=24 x \quad \text{... (iii)}\\ \end{aligned}$

From equation (i), (ii) and (iii)

$\begin{aligned} \Rightarrow & \frac{1}{y(x)}\left[\left(x^2-1\right) \frac{d^2 y(x)}{d x^2}+x \frac{d y(x)}{d x}\right] \\ & =\frac{1}{4 x^3-3 x}\left[\left(x^2-1\right) \cdot 24 x+x \cdot\left(12 x^2-3\right)\right] \\ & =\frac{1}{4 x^3-3 x}\left[24 x^3-24 x+12 x^3-3 x\right] \\ & =\frac{1}{4 x^3-3 x}\left[36 x^3-27 x\right] \end{aligned}$

$\begin{aligned} & =\frac{9\left(4 x^3-3 x\right)}{4 x^3-3 x} \\ & =9 \end{aligned}$

Hence P match with 4

(Q) Given $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3, \ldots . \mathrm{A}_n(n>2)$ be the vertices of a regular polygon of $n$ sides with centre at origin and $\vec{a}_k$ be the position vector of the point $A_k$

$\begin{aligned} & \text { Here, } \begin{aligned} &\left|\vec{a}_1\right|=\left|\vec{a}_2\right|=\left|\vec{a}_3\right|=\ldots=\left|\vec{a}_k\right| \\ &=\left|\vec{a}_{k+1}\right|=\cdots=\left|\vec{a}_n\right|=\lambda \text { (let) } \\ & \text { And } \angle \mathrm{A}_1 \mathrm{OA}_2=\angle \mathrm{A}_2 \mathrm{OA}_3=\cdots=\angle \mathrm{A}_k \mathrm{OA}_{k+1} \\ &=\cdots=\angle \mathrm{A}_n \mathrm{OA}_1=\frac{2 \pi}{n} \\ & \Rightarrow \vec{a}_k \times \vec{a}_{k+1}=\lambda \cdot \lambda \cdot \sin \frac{2 \pi}{n} \cdot \hat{p} \text { and } \\ & \vec{a}_k \cdot \vec{a}_{k+1}=\lambda \cdot \lambda \cdot \cos \frac{2 \pi}{n} \end{aligned} \end{aligned}$

Apply $\left|\sum_\limits{k=1}^{n-1} \vec{a}_k \times \vec{a}_{k+1}\right|=\left|\sum_\limits{k=1}^{n-1} \vec{a}_k \cdot \vec{a}_{k+1}\right|$

$\begin{aligned} & \left|\sum_{k=1}^{n-1} \lambda^2 \sin \frac{2 \pi}{n} \hat{p}\right|=\left|\sum_{k=1}^{n-1} \lambda^2 \cos \frac{2 \pi}{n}\right| \\ & \Rightarrow(n-1) \lambda^2\left|\sin \frac{2 \pi}{n}\right|=(n-1) \lambda^2\left|\cos \frac{2 \pi}{n}\right| \\ & \Rightarrow\left|\tan \frac{2 \pi}{n}\right|=1 \\ & \Rightarrow n_{\text {min }}=8 \\ \end{aligned}$

Hence $Q$ match with 3 .

(R) We know the equation of normal of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $\left(x_1, y_1\right)$ is $\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2$.

$\therefore$ Equation of Normal of ellipse $\frac{x^2}{6}+\frac{y^2}{3}=1$ at $p(h, 1)$ is $\frac{6 x}{h}-\frac{3 y}{1}=1$

Given, the normal at $p(h, 1)$ is perpendicular to the line $x+y=3$

$\Rightarrow \quad \frac{{\frac{6}{h}}}{3} \times(-1)=-1$

$\Rightarrow \quad \frac{2}{h}=1$

$\Rightarrow \quad h=2$

Hence, R match with 2.

$\begin{aligned} & \text { (S) } \tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1} \frac{2}{x^2} \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{\frac{1}{2 x+1}+\frac{1}{4 x+1}}{1-\frac{1}{2 x+1} \times \frac{1}{4 x+1}}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{6 x+2}{8 x^2+6 x+1-1}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right) \\ & \Rightarrow \quad \tan ^{-1}\left(\frac{3 x+1}{4 x^2+3 x}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right) \end{aligned}$

$\begin{aligned} & \Rightarrow & 3 x^3+x^2 & =8 x^2+6 x \\ & \Rightarrow & x\left[3 x^2-7 x-6\right] & =0 \\ & \Rightarrow & x\left[3 x^2-9 x+2 x-6\right] & =0 \\ & \Rightarrow & x(3 x+2)(x-3) & =0 \\ & \Rightarrow & x & =0, \frac{-2}{3}, 3 \end{aligned}$

So, $x=3$ is the only positive solution of given equation.

Hence, S match with 1.

Hint:

(i) Recall $\cos \left(\cos ^{-1} x\right)=x \forall x \in[-1,1]$ and $\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta$

(ii) If angle between $\vec{a}$ and $\vec{b}$ is $\theta$, then $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$ and $\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}$

(iii) The equation of normal of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $\left(x_1, y_1\right)$ is $\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2$.

(iv) Recall $\tan ^{-1} \mathrm{~A}+\tan ^{-1} \mathrm{~B}=\tan ^{-1}\left(\frac{\mathrm{A}+\mathrm{B}}{1-\mathrm{AB}}\right)$ only when A.B < 1

For P

$\text { Let } t=\sqrt{\frac{1}{y^2}\left(\frac{\cos \left(\tan ^{-1} y\right)+y \cdot \sin \left(\tan ^{-1} y\right)}{\cot \left(\sin ^{-1} y\right)+\tan \left(\sin ^{-1} y\right)}\right)^2+y^4}$

We know that

$\begin{gathered} \tan ^{-1} y=\cos ^{-1}\left(\frac{1}{\sqrt{1+y^2}}\right)=\sin ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) \\ \text { and } \sin ^{-1} y=\cot ^{-1}\left(\frac{\sqrt{1-y^2}}{y}\right)=\tan ^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right) \end{gathered}$

$\therefore \quad t = \sqrt {{1 \over {{y^2}}}{{\left[ {{{\cos \left( {{{\cos }^{ - 1}}{1 \over {\sqrt {1 + {y^2}} }}} \right) + y\,.\,\sin \left( {{{\sin }^{ - 1}}{y \over {\sqrt {1 + {y^2}} }}} \right)} \over {\cot \left( {{{\cot }^{ - 1}}{{\sqrt {1 - {y^2}} } \over y}} \right) + \tan \left( {{{\tan }^{ - 1}}{y \over {\sqrt {1 - {y^2}} }}} \right)}}} \right]}^2} + {y^4}} $

$\begin{aligned} & \Rightarrow t=\sqrt{\frac{1}{y^2} \times\left[\frac{\frac{1}{\sqrt{1+y^2}}+\frac{y^2}{\sqrt{1+y^2}}}{\frac{\sqrt{1-y^2}}{y}+\frac{y}{\sqrt{1-y^2}}}\right]^2+y^4} \\ & \Rightarrow t=\sqrt{\frac{1}{y^2} \times\left[\frac{\sqrt{1+y^2}}{\frac{1-y^2+y^2}{y \sqrt{1-y^2}}}\right]^2+y^4} \\ & \Rightarrow t=\sqrt{\frac{1}{y^2} \times y^2\left(1+y^2\right)\left(1-y^2\right)+y^4} \\ & \Rightarrow t=\sqrt{\left(1-y^4\right)+y^4}=1 \\ \end{aligned}$

Hence, P match with 4.

For Q.

Given $\cos x+\cos y+\cos z=0$

$\Rightarrow \cos x+\cos y=-\cos z$

On squaring both side

$\Rightarrow \cos ^2 x+\cos ^2 y+2 \cos x \cdot \cos y=\cos ^2 z\quad \text{... (i)}$

Also given $\sin x+\sin y+\sin z=0$

$\Rightarrow \sin x+\sin y=-\sin z$

On squaring both side

$\Rightarrow \sin ^2 x+\sin ^2 y+2 \sin x \cdot \sin y=\sin ^2 z \quad \text{... (ii)}$

Add equation (i) and (ii)

$\begin{gathered} \Rightarrow\left(\cos ^2 x+\sin ^2 x\right)+\left(\cos ^2 y+\sin ^2 y\right) \\ +2[\cos x \cdot \cos y+\sin x \cdot \sin y] \\ =\cos ^2 z+\sin ^2 z \end{gathered}$

$\begin{array}{lr} \Rightarrow & 1+1+2 \cos (x-y)=1 \\ \Rightarrow & \cos (x-y)=\frac{-1}{2} \\ \Rightarrow & 2 \cos ^2\left(\frac{x-y}{2}\right)-1=\frac{-1}{2} \end{array}$

$\begin{array}{ll} \Rightarrow & \cos ^2\left(\frac{x-y}{2}\right)=\frac{1}{4} \\ \Rightarrow & \cos \left(\frac{x-y}{2}\right)= \pm \frac{1}{2} \end{array}$

Hence, Q match with 3.

For R

Given,

$\begin{aligned} & \cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \cdot \sin 2 x \cdot \sec x \\ & =\cos x \cdot \sin 2 x \cdot \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x \end{aligned}$

$\begin{aligned} \Rightarrow & {\left[\cos \left(\frac{\pi}{4}-x\right)-\cos \left(\frac{\pi}{4}+x\right)\right] } \\ & \cos 2 x+(\sin x-\cos x) \sin 2 x \cdot \sec x=0 \end{aligned}$

$\begin{aligned} \Rightarrow & 2 \cdot \sin \frac{\pi}{4} \cdot \sin x \cdot \cos 2 x+(\sin x-\cos x) \\ & \frac{2 \sin x \cdot \cos x}{\cos x}=0 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad \sqrt{2} \cos 2 x=2(\cos x-\sin x) \\ & \Rightarrow \quad\left(\cos ^2 x-\sin ^2 x\right)=\sqrt{2} (\cos x-\sin x) \\ \end{aligned}$

$\begin{array}{rlrl} \Rightarrow & \cos x+\sin x =\sqrt{2} \\ \Rightarrow & (\cos x+\sin x)^2 =(\sqrt{2})^2 \\ \Rightarrow & \cos ^2 x+\sin ^2 x+2 \sin x \cdot \cos x =2 \\ \Rightarrow & 1+\sin 2 x =2 \\ \Rightarrow & \sin 2 x =1 \\ \Rightarrow & 2 x =\frac{\pi}{2} \\ \Rightarrow & x =\frac{\pi}{4} \\ \Rightarrow & \sec x =\sec \frac{\pi}{4}=\sqrt{2} \end{array}$

Hence, R match with 2

For S

Given,

$\begin{aligned} \cot \left(\sin ^{-1} \sqrt{1-x^2}\right) & =\sin \left(\tan ^{-1}(x \sqrt{6})\right), x \neq 0 \\ \Rightarrow \cot \left(\cot ^{-1} \frac{x}{\sqrt{1-x^2}}\right) & =\sin \left(\sin ^{-1} \frac{x \sqrt{6}}{\sqrt{1+6 x^2}}\right) \\ \Rightarrow \quad \frac{x}{\sqrt{1-x^2}} & =\frac{x \sqrt{6}}{\sqrt{1+6 x^2}} \end{aligned}$

$ \begin{array}{rlrl} \Rightarrow & \frac{1}{\sqrt{1-x^2}} & =\frac{\sqrt{6}}{\sqrt{1+6 x^2}} \\ \Rightarrow & \text { on squaring both sides } \\ \Rightarrow & \frac{1}{1-x^2} =\frac{6}{1+6 x^2} \\ \Rightarrow & 1+6 x^2=6-6 x^2 \\ \Rightarrow & x^2 =\frac{5}{12} \\ \Rightarrow & x= \pm \sqrt{\frac{5}{12}}= \pm \frac{1}{2} \sqrt{\frac{5}{3}} \end{array}$

Hence, S match with 1.

Hints :

(i) Recall the method of conversion of one inverse trigonometric function into other inverse trigonometric function as

$\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^2}=\tan ^{-1} \frac{x}{\sqrt{1-x^2}}$

(ii) Recall $\cos \left(\cos ^{-1} x\right)=x, \sin \left(\sin ^{-1} x\right)=x\tan \left(\tan ^{-1} x\right)=x \text { etc }$

(iii) Recall the following trigonometric identities

(A) $\sin ^2 x+\cos ^2 x=1$

(B) $\sin 2 x=2 \sin x \cdot \cos x$

(C) $\cos 2 x=\cos ^2 x-\sin ^2 x=2 \cos ^2 x-1= 1-2 \sin ^2 x$

(D) $\cos (A-B)-\cos (A+B)=2 \sin A \cdot \sin B$

$\cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + 2 \times {{n(n + 1)} \over 2}} \right)} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}(1 + n(n + 1))} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {{{n(n + 1) + 1} \over {(n + 1) - n}}} \right)} } \right)$

$ = \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}n - {{\cot }^{ - 1}}(n + 1)} } \right)$

$ = \cot (({\cot ^{ - 1}}1 + {\cot ^{ - 1}}2 + {\cot ^{ - 1}}3 + .... + {\cot ^{ - 1}}23) - ({\cot ^{ - 1}}2 + {\cot ^{ - 1}}3 + .... + {\cot ^{ - 1}}23 + {\cot ^{ - 1}}24))$

$ = \cot ({\cot ^{ - 1}}1 - {\cot ^{ - 1}}24)$

$ = \cot \left( {{{\cot }^{ - 1}}{{24 \times 1 + 1} \over {24 - 1}}} \right) = {{25} \over {23}}$

Given that,

$0 < x < 1$,

$\sqrt {1 + {x^2}} {[{\{ x\cos ({\cot ^{ - 1}}x) + \sin ({\cot ^{ - 1}}x)\} ^2} - 1]^{{1 \over 2}}}$

Find : value of this expression

Here, $0 < x < 1$,

${\cot ^{ - 1}}x = {\sin ^{ - 1}}\left( {{1 \over {\sqrt {1 + {x^2}} }}} \right) = {\cos ^{ - 1}}\left( {{x \over {\sqrt {1 + {x^2}} }}} \right)$

Now, $\sqrt {1 + {x^2}} {[{\{ x\cos ({\cot ^{ - 1}}x) + \sin ({\cot ^{ - 1}}x)\} ^2} - 1]^{{1 \over 2}}}$

$\sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cos }^{ - 1}}\left( {{x \over {1 + {x^2}}}} \right)} \right) + \sin \left( {{{\sin }^{ - 1}}{1 \over {\sqrt {1 + {x^2}} }}} \right)} \right\}}^2}} \right]^{{1 \over 2}}}$

$ = \sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\,.\,{x \over {\sqrt {1 + {x^2}} }} + {1 \over {\sqrt {1 + {x^2}} }}} \right\}}^2} - 1} \right]^{{1 \over 2}}}$

$ = \sqrt {1 + {x^2}} {\left[ {{{\left\{ {{{{x^2} + 1} \over {\sqrt {1 + {x^2}} }}} \right\}}^2} - 1} \right]^{{1 \over 2}}}$

$ = \sqrt {1 + {x^2}} {[{x^2} + 1 - 1]^{{1 \over 2}}}$

$ = x\sqrt {1 + {x^2}} $

which is the value of the expression.

(A) If $a=1$ and $b=0$ then

$\begin{aligned} & \sin ^{-1} x+\cos ^{-1} y+\frac{\pi}{2}=\frac{\pi}{2} \\ & \sin ^{-1} x=-\cos ^{-1} y \\ & \sin ^{-1} x=-\sin ^{-1} \sqrt{1-y^{2}} \\ &-x=\sqrt{1-y^{2}} \\ & x^{2}+y^{2}=1 \\ & \text { (A) } \rightarrow(p) \end{aligned}$

(B) If $a=1$ and $b=1$ then

$\begin{aligned} & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1} x y=\frac{\pi}{2} \\ & \cos ^{-1} x-\cos ^{-1} y=\cos ^{-1} x y \\ & x y+\sqrt{1-y^{2}} \sqrt{1-x^{2}}=x y \\ & =\left(x^{2}-1\right)\left(y^{2}-1\right)=0 \\ & (\mathrm{~B}) \rightarrow(q) \end{aligned}$

(C) If $a=1$ and $b=2$ then

$\begin{aligned} & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1}(2 x y) \\ & =-\cos ^{-1} x-\cos ^{-1} y=\cos ^{-1}(2 x y) \\ & \quad x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}=2 x y \end{aligned}$

$\begin{gathered} \sqrt{1-x^{2}} \sqrt{1-y^{2}}=x y \\ 1-x^{2}-y^{2}+x^{2} y^{2}=x^{2} y^{2} \\ x^{2}+y^{2}=1 \\ (\mathrm{C}) \rightarrow(p) \end{gathered}$

(D) If $a=2$ and $b=2$, we get

$\begin{gathered} \sin ^{-1} 2 x+\cos ^{-1}(2 x y)+\cos ^{-1} y=\frac{\pi}{2} \\ \cos ^{-1}(2 x)-\cos ^{-1} y=\cos ^{-1}(2 x y) \\ 2 x y+\sqrt{4 x^{2}-1} \sqrt{1-y^{2}}=2 x y \\ \left(4 x^{2}-1\right)\left(y^{2}-1\right)=0 \\ \text { (D) } \rightarrow(\mathrm{s}) \end{gathered}$

$f(x) = \int {{{\sin }^2}xdx = {1 \over 2}\int {(1 - \cos 2x)dx} } $

$ = {1 \over 4}(2x - \sin 2x) + c$

Now,

$f(x + \pi ) = {1 \over 4}(2x + 2\pi - \sin (2x + 2\pi )) + c$

$ = {1 \over 4}[2x + 2\pi - \sin 2x] + c \ne f(x)$

$\therefore$ Statement 1 is false.

Also ${\sin ^2}(x + \pi ) = {\sin ^2}x,\forall x \in R$

$\therefore$ Statement 2 is true.

To determine the principal value of $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$, we need to understand the properties of the inverse sine function, also known as the arcsine function.

The sine function, $\sin(x)$, is periodic with period $2\pi$, meaning that for any integer $k$, we have:

$\sin(x) = \sin(x + 2k\pi)$.

The inverse sine function, $\sin^{-1}(x)$, has a restricted domain, typically taken as:

$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$,

where $y = \sin^{-1}(x)$.

Given the expression inside the inverse sine function:

$\sin\left(\frac{2\pi}{3}\right)$,

we know:

$\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

Now, we want to find the principal value of $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$.

Let:

$x = \frac{2\pi}{3}$,

so:

$\sin\left(\frac{2\pi}{3}\right) = \sin(x) = \frac{\sqrt{3}}{2}$.

To find the principal value, we need to find an angle $y$ within the range $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$ such that $\sin(y) = \frac{\sqrt{3}}{2}$.

The principal value that satisfies this condition is:

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

Given the range of the inverse sine function, the corresponding angle is:

$y = \frac{\pi}{3}$.

Now, we need to match this value with one of the given options. Since there is no option for $\frac{\pi}{3}$ directly, but recalling the properties of the sine function and the range, and knowing that $\frac{2\pi}{3}$ is not within the range of the principal values, we need to reconsider equivalent values.

None of the given options match exactly the principal value we calculated, which is $\frac{\pi}{3}$. Therefore, the correct answer is:

Option D: none

Here, $\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$

and $\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$ as ${6 \over {11}} > {1 \over 2}$

$ \Rightarrow {\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\sin ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 6}$

$\therefore$ $\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\pi \over 2} \Rightarrow \cos \alpha < 0$

Now, $\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$

As ${4 \over 9} < {1 \over 2} \Rightarrow {\cos ^{ - 1}}\left( {{4 \over 9}} \right) > {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}$

$\therefore$ $\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right) > \pi $

$\therefore$ $\cos \beta < 0$ and $\sin \beta < 0$

Now, $\alpha + \beta $ is slightly greater than ${{3\pi } \over 2}$.

$\therefore$ $\cos (\alpha + \beta ) > 0$