Question

The expression ${\sin ^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}\left( x \right)} \right)} \right) + {\cos ^{ - 1}}\left( {\sin \left( {{{\cos }^{ - 1}}\left( x \right)} \right)} \right)$ is equal to
(a) $\dfrac{\pi }{2}$
(b) $\dfrac{\pi }{4}$
(c) $\dfrac{{3\pi }}{4}$
(d) 0

Answer 1

Here, we need to find the value of the given expression. We will find the value of ${\sin ^{ - 1}}\left( x \right)$ in terms of cosine inverse of an angle. Similarly, we will find the value of ${\cos ^{ - 1}}\left( x \right)$ in terms of sine inverse of an angle. Then, we will use the obtained equations to simplify the given expression. Finally, we will use the identities for trigonometric inverse functions to simplify the expression further, and get the required answer.
Formula Used: We will use the following formulas:

1. The sine of an angle $\theta$ in a right angled triangle is given by $\sin \theta = \dfrac{P}{H}$, where $P$ is the perpendicular and $H$ is the hypotenuse.
2. The cosine of an angle $\theta$ in a right angled triangle is given by $\cos \theta = \dfrac{B}{H}$, where $B$ is the base and $H$ is the hypotenuse.
3. The identities $\cos \left( {{{\cos }^{ - 1}}x} \right) = x$ and $\sin \left( {{{\sin }^{ - 1}}x} \right) = x$.
4. The sum of ${\sin ^{ - 1}}x$ and ${\cos ^{ - 1}}x$ is equal to $\dfrac{\pi }{2}$, if $\left| x \right| < 1$, that is ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$.

Complete step by step solution:
Let ${\sin ^{ - 1}}x = \phi$.
Rewriting the expression, we get
$\sin \phi = x$
We know that the sine of an angle $\theta$ in a right angled triangle is given by $\sin \theta = \dfrac{P}{H}$.
From the equations $\sin \phi = x$ and $\sin \phi = \dfrac{P}{H}$, we get
$\begin{array}{l} \Rightarrow x = \dfrac{P}{H}\\\ \Rightarrow \dfrac{x}{1} = \dfrac{P}{H}\end{array}$
Therefore, we get perpendicular $= x$ and hypotenuse $= 1$.
Now, the Pythagoras’s theorem states that the square of the hypotenuse is equal to the sum of the squares of the base and the perpendicular.
Thus, we get
${H^2} = {B^2} + {P^2}$
Here, $B$ is the base.
Substituting perpendicular $= x$ and hypotenuse $= 1$ in the formula, we get
$\begin{array}{l} \Rightarrow {1^2} = {B^2} + {x^2}\\\ \Rightarrow 1 = {B^2} + {x^2}\end{array}$
Rewriting the equation, we get
$\Rightarrow {B^2} = 1 - {x^2}$
Taking the square root on both the sides, we get
$\Rightarrow B = \sqrt {1 - {x^2}}$
Now, the cosine of an angle $\theta$ in a right angled triangle is given by $\cos \theta = \dfrac{B}{H}$, where $B$ is the base and $H$ is the hypotenuse.
Thus, we get
$\begin{array}{l} \Rightarrow \cos \phi = \dfrac{{\sqrt {1 - {x^2}} }}{1}\\\ \Rightarrow \cos \phi = \sqrt {1 - {x^2}} \end{array}$
Rewriting the equation, we get
$\Rightarrow {\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) = \phi$
From the equations ${\sin ^{ - 1}}x = \phi$ and ${\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) = \phi$, we get
$\Rightarrow {\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$
Now, let ${\cos ^{ - 1}}x = \theta$.
Rewriting the expression, we get
$\cos \theta = x$
Now we know that the cosine of an angle $\theta$ in a right angled triangle is given by $\cos \theta = \dfrac{B}{H}$.
From the equations $\cos \theta = x$ and $\cos \theta = \dfrac{B}{H}$, we get
$\begin{array}{l} \Rightarrow x = \dfrac{B}{H}\\\ \Rightarrow \dfrac{x}{1} = \dfrac{B}{H}\end{array}$
Therefore, we get base $= x$ and hypotenuse $= 1$.
From Pythagoras’s theorem, we have
${H^2} = {B^2} + {P^2}$
Substituting base $= x$ and hypotenuse $= 1$ in the formula, we get
$\begin{array}{l} \Rightarrow {1^2} = {x^2} + {P^2}\\\ \Rightarrow 1 = {x^2} + {P^2}\end{array}$
Rewriting the equation, we get
$\Rightarrow {P^2} = 1 - {x^2}$
Taking the square root on both the sides, we get
$\Rightarrow P = \sqrt {1 - {x^2}}$
Using the formula $\sin \theta = \dfrac{P}{H}$, we get
$\begin{array}{l} \Rightarrow \sin \theta = \dfrac{{\sqrt {1 - {x^2}} }}{1}\\\ \Rightarrow \sin \theta = \sqrt {1 - {x^2}} \end{array}$
Rewriting the equation, we get
$\Rightarrow {\sin ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) = \theta$
From the equations ${\cos ^{ - 1}}x = \theta$ and ${\sin ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) = \theta$, we get
$\Rightarrow {\cos ^{ - 1}}x = {\sin ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$
Now, we will simplify the given expression.
Substituting ${\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$ and ${\cos ^{ - 1}}x = {\sin ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$ in the expression ${\sin ^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}\left( x \right)} \right)} \right) + {\cos ^{ - 1}}\left( {\sin \left( {{{\cos }^{ - 1}}\left( x \right)} \right)} \right)$, we get
$\begin{array}{l} \Rightarrow {\sin ^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}\left( x \right)} \right)} \right) + {\cos ^{ - 1}}\left( {\sin \left( {{{\cos }^{ - 1}}\left( x \right)} \right)} \right)\\\ = {\sin ^{ - 1}}\left( {\cos \left( {{{\cos }^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)} \right)} \right) + {\cos ^{ - 1}}\left( {\sin \left( {{{\sin }^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)} \right)} \right)\end{array}$
We know that $\cos \left( {{{\cos }^{ - 1}}x} \right) = x$ and $\sin \left( {{{\sin }^{ - 1}}x} \right) = x$.
Thus, we get
$\Rightarrow {\sin ^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}\left( x \right)} \right)} \right) + {\cos ^{ - 1}}\left( {\sin \left( {{{\cos }^{ - 1}}\left( x \right)} \right)} \right) = {\sin ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) + {\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$
The sum of ${\sin ^{ - 1}}x$ and ${\cos ^{ - 1}}x$ is equal to $\dfrac{\pi }{2}$, if $\left| x \right| < 1$, that is ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$.
Therefore, we get
$\Rightarrow {\sin ^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}\left( x \right)} \right)} \right) + {\cos ^{ - 1}}\left( {\sin \left( {{{\cos }^{ - 1}}\left( x \right)} \right)} \right) = \dfrac{\pi }{2}$
Thus, the value of the given expression is $\dfrac{\pi }{2}$.

$\therefore$ The correct option is option (a).

Note:
We can apply the identity ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$ only if $\left| x \right| < 1$.
We will verify that $\left| {\sqrt {1 - {x^2}} } \right|$ is less than 1.
We know that the square of a number is always greater than 0.
Thus, we get
${x^2} > 0$
The sign of inequality changes if both sides are multiplied by the same negative number.
Multiplying both sides by $- 1$, we get
$\begin{array}{l} \Rightarrow {x^2} \times \left( { - 1} \right) < \- 0 \times \left( { - 1} \right)\\\ \Rightarrow - {x^2} < 0\end{array}$
Adding 1 on both sides, we get
$\begin{array}{l} \Rightarrow - {x^2} + 1 < 0 + 1\\\ \Rightarrow 1 - {x^2} < 1\end{array}$
Taking the square root on both the sides, we get
$\begin{array}{l} \Rightarrow \sqrt {1 - {x^2}} < \sqrt 1 \\\ \Rightarrow \sqrt {1 - {x^2}} < 1\\\ \Rightarrow \left| {\sqrt {1 - {x^2}} } \right| < 1\end{array}$
Hence, we have proved that $\left| {\sqrt {1 - {x^2}} } \right| < 1$.