Inverse Trigonometric Functions

Given sin^-1$\left( {{x \over 5}} \right)$ + cosec^-1$\left( {{5 \over 4}} \right)$ = ${\pi \over 2}$

$ \Rightarrow $ sin^-1$\left( {{x \over 5}} \right)$ + sin^-1$\left( {{4 \over 5}} \right)$ = ${\pi \over 2}$

$ \Rightarrow $ sin^-1$\left( {{x \over 5}} \right)$ = ${\pi \over 2}$ - sin^-1$\left( {{4 \over 5}} \right)$

$ \Rightarrow $ sin^-1$\left( {{x \over 5}} \right)$ = cos^-1$\left( {{4 \over 5}} \right)$

$ \Rightarrow $ ${x \over 5}$ = sin(cos^-1$ {{4 \over 5}}$)

$ \Rightarrow $ ${x \over 5}$ = sin(sin^-1$ {{3 \over 5}}$)

$ \Rightarrow $ ${x \over 5}$ = ${3 \over 5}$

$ \Rightarrow $ x = 3

(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.

As we know,

${\cos ^{ - 1}}A - {\cos ^{ - 1}}B$

$ = {\cos ^{ - 1}}\left( {AB + \sqrt {1 - {A^2}} .\sqrt {1 - {B^2}} } \right)$

Given, ${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $

$ \Rightarrow {\cos ^{ - 1}}\left( {x.{y \over 2} + \sqrt {1 - {x^2}} .\sqrt {1 - {{{y^2}} \over 4}} } \right) = \alpha $

$ \Rightarrow {{xy} \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{y{}^2} \over 4}} = \cos \,x$

$ \Rightarrow {\left( {\cos x - {{xy} \over 2}} \right)^2} = \left( {1 - {x^2}} \right)\left( {1 - {{{y^2}} \over 4}} \right)$

$ \Rightarrow {\cos ^2} + {{{x^2}{y^2}} \over 4} - 2.\cos x.{{xy} \over 2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - {x^2} - {{{y^2}} \over 4} + {{{x^2}{y^2}} \over 4}$

$ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\,\cos x = 1 - {\cos ^2}x$

$ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\cos x = {\sin ^2}x$

$ \Rightarrow 4{x^2} + y{}^2 - 4xy\cos x = 4{\sin ^2}x$

Given that,

${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$

We know,

$ - {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}$

$\therefore$ $\,\,\,$ $ - {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}$

$ \Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over 4}$

$ \Rightarrow \sin \left( { - {\pi \over 4}} \right) \le a \le \sin \left( {{\pi \over 4}} \right)$

$ \Rightarrow - {1 \over {\sqrt 2 }} \le a \le {1 \over {\sqrt 2 }}$

$\therefore$ $\,\,\,\,\,\,\,\,\left| a \right| \le {1 \over {\sqrt 2 }}$

Given that,

${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) - {\tan ^{ - 1}}\left( \sqrt{\cos x} \right) = x\,\,\,\,...\left( 1 \right)$

We know,

${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = {\pi \over 2}$

$\therefore$ $\,\,\,$ ${\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) + {\tan ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {\pi \over 2}\,\,\,...\left( 2 \right)$

Adding $(1)$ and $(2),$ we get,

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

$ \Rightarrow \,\,{\cot ^{ - 1}}\left( {\sqrt {\cos x} } \right) = {x \over 2} + {\pi \over 4}$

$ \Rightarrow \,\,\sqrt {\cos x} = \cot \left( {{x \over 2} + {\pi \over 4}} \right)$

$ \Rightarrow \,\,\sqrt {\cos x} = {{\cot {x \over 2} - 1} \over {1 + \cot {x \over 2}}}$

$ \Rightarrow \,\,\sqrt {\cos x} = {{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}}$

Squaring both sides we get,

$ \Rightarrow \,\,\cos x = {{1 - 2\sin {x \over 2}\cos {x \over 2}} \over {1 + 2\sin {x \over 2}\cos {x \over 2}}}$

$ \Rightarrow \,\,\cos x = {{1 - \sin x} \over {1 + \sin x}}$

$ \Rightarrow \,\,{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}} = {{1 - \sin x} \over {1 + \sin x}}$

Applying compounds and dividendo rule,

$ \Rightarrow \,\,{{2\sin x} \over 2} = {{2{{\tan }^2}{x \over 2}} \over 2}$

$ \Rightarrow \,\,\sin x = {\tan ^2}{x \over 2}$

Other Method :

$ \begin{aligned} & \text { Since, } \cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \\\\ & \Rightarrow \tan ^{-1}\left(\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}\right)=x \\\\ & \Rightarrow \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=\tan x \\\\ & \Rightarrow \cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha} \\\\ & \because \operatorname{cosec} x=\sqrt{1+\cot { }^2 x} \\\\ & \therefore \operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha} \\\\ & \Rightarrow \sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\tan ^2 \frac{\alpha}{2} \end{aligned} $

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