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If ${{\cos }^{-1}}x+{{\cos }^{-1}}y=\dfrac{\pi }{2}$ , then prove that ${{\cos }^{-1}}x={{\sin }^{-1}}y$

Last updated date: 10th Aug 2024
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Hint: In the problem, we have the inverse trigonometric functions, so we will assume each inverse trigonometric ratio as a variable and calculates the relationship between the variables by using the given conditions. Now we will calculate the value of ${{\sin }^{-1}}y$ by using all the values we have calculated so far.

Given that,
${{\cos }^{-1}}x+{{\cos }^{-1}}y=\dfrac{\pi }{2}$
Let ${{\cos }^{-1}}x=a\Rightarrow x=\cos a$
Let ${{\cos }^{-1}}y=b\Rightarrow y=\cos b$
Applying the above values in the given equation, then we will get
$a+b=\dfrac{\pi }{2}...\left( \text{i} \right)$
We have to prove that ${{\cos }^{-1}}x={{\sin }^{-1}}y$ . For this we will calculate the value of ${{\sin }^{-1}}y$ .
We have $y=\cos b$ .
From equation $\left( \text{i} \right)$ we have the value of $b=\dfrac{\pi }{2}-a$ , substituting the value of $b$ in the value of $y$ . Then we will get
\begin{align} & y=\cos b \\ & \Rightarrow y=\cos \left( \dfrac{\pi }{2}-a \right) \\ \end{align}
We know the $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta$ , then we will get
$\Rightarrow y=\sin a$
Applying the ${{\sin }^{-1}}$ function in the above equation, then we will get
$\Rightarrow {{\sin }^{-1}}\left( y \right)={{\sin }^{-1}}\left( \sin a \right)$
We know that ${{\sin }^{-1}}\left( \sin \theta \right)=\theta$ , then we will get
$\Rightarrow {{\sin }^{-1}}y=a$
But we have considered the value $a={{\cos }^{-1}}x$ , then we will get
${{\cos }^{-1}}x={{\sin }^{-1}}y$ .
Hence proved.

Note:
For this problem we can also prove that ${{\cos }^{-1}}y={{\sin }^{-1}}x$ by calculating the value of ${{\sin }^{-1}}x$ .
From equation $\left( \text{i} \right)$ we have the value of $a=\dfrac{\pi }{2}-b$ , substituting the value of $a$ in the value of $x$ . Then we will get
\begin{align} & x=\cos a \\ & \Rightarrow x=\cos \left( \dfrac{\pi }{2}-b \right) \\ \end{align}
We know the $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta$ , then we will get
$\Rightarrow x=\sin b$
Applying the ${{\sin }^{-1}}$ function in the above equation, then we will get
$\Rightarrow {{\sin }^{-1}}\left( x \right)={{\sin }^{-1}}\left( \sin b \right)$
We know that ${{\sin }^{-1}}\left( \sin \theta \right)=\theta$ , then we will get
$\Rightarrow {{\sin }^{-1}}x=b$
But we have considered the value $b={{\cos }^{-1}}y$ , then we will get
${{\cos }^{-1}}y={{\sin }^{-1}}x$