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How do you prove \[{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}\]?

Last updated date: 25th Feb 2024
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Hint: To solve the given question, we should know some of the trigonometric properties of sine and cosine function. We should know that \[\sin \left( \dfrac{\pi }{2}-x \right)=\cos x\]. In other words, if the sine of an angle is equal to the cosine of another angle, then the two angles should be complementary. We will use this property to solve this question.

Complete step-by-step solution:
We are asked to prove the statement. To prove a statement, we need to show that either the left-hand side or right-hand side of the given statement can be simplified to express the other side. Here the left-hand side and right-hand side of the statement are \[{{\sin }^{-1}}x+{{\cos }^{-1}}x\]and \[\dfrac{\pi }{2}\] respectively. Let’s take the LHS to simplify, assume that \[{{\sin }^{-1}}x=A\], and \[{{\cos }^{-1}}x=B\]. Taking sine of both sides of the first equation, and cosine of both sides of the second equation, we get \[\sin A=x\And \cos B=x\]. Thus, we can say that \[\sin A=\cos B\]. Using the property that if the sine of angle is equal to cosine of another angle, then the two angles should be complementary. We can say that
\[\Rightarrow A+B=\dfrac{\pi }{2}\]
From the substitution, we have \[{{\sin }^{-1}}x=A\], and \[{{\cos }^{-1}}x=B\]. Thus, the above expression can be written as
\[\Rightarrow {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}=RHS\]
\[\therefore LHS=RHS\]
Hence proved.

Note: The statement given in the above question is one of the properties of inverse trigonometric functions, so it should be remembered. We can also use similar property for other inverse trigonometric functions as, \[{{\csc }^{-1}}x+{{\sec }^{-1}}x=\dfrac{\pi }{2}\] and \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\].
These properties are useful to solve other questions based on evaluation of expression, and proofs.