Question

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

Answer 1

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.

Complete step by step answer:
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 $
So, they may ask about both the conditions.