Question

How do you prove that $\sin \left( {{\cot }^{-1}}\left( \tan \left( {{\cos }^{-1}}\left( \sin x \right) \right) \right) \right)=\sin x$ ?

Answer 1

Hint: In this question, we have to prove a trigonometric equation. Thus, we will apply the trigonometric formulas to get the solution. First, we will change the sin function into cos function, by using the angle formula, after that we will apply the inverse-trigonometric formula \[{{\cos }^{-1}}(\cos x)=x\] . Then, we will again convert the tan function into cot function using the angle formula. In the end we will apply the inverse-trigonometric formula \[{{\cot }^{-1}}(\cot x)=x\] , which is equal to the right-hand side of the equation, which is the required solution to the problem.

Complete step by step solution:
According to the question, we have to prove the trigonometric equation.
Thus, we will use trigonometric formulas to get the solution.
The trigonometric equation given to us is $\sin \left( {{\cot }^{-1}}\left( \tan \left( {{\cos }^{-1}}\left( \sin x \right) \right) \right) \right)=\sin x$ ----- (1)
So, we start solving this problem by using the left-hand side of the equation (1), we get
$LHS=\sin \left( {{\cot }^{-1}}\left( \tan \left( {{\cos }^{-1}}\left( \sin x \right) \right) \right) \right)$
Let us first solve the brackets of the above equation, by converting the sin function into cos function using the angle formula $\cos \left( \dfrac{\pi }{2}-x \right)=\sin x$ , we get
$\Rightarrow \sin \left( {{\cot }^{-1}}\left( \tan \left( {{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{2}-x \right) \right) \right) \right) \right)$
Now, we will again use the trigonometric formula \[{{\cos }^{-1}}(\cos x)=x\] in the above equation, we get
$\Rightarrow \sin \left( {{\cot }^{-1}}\left( \tan \left( \dfrac{\pi }{2}-x \right) \right) \right)$
So, again we will convert the tan function into cos function using the angle formula $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$ , we get
$\Rightarrow \sin \left( {{\cot }^{-1}}\left( \cot x \right) \right)$
Now, we will again use the trigonometric formula \[{{\cot }^{-1}}(\cot x)=x\] in the above equation, we get
$\Rightarrow \sin x=RHS$
Therefore, we proved that the left-hand side is equal to the right-hand side, that is $\sin \left( {{\cot }^{-1}}\left( \tan \left( {{\cos }^{-1}}\left( \sin x \right) \right) \right) \right)=\sin x$ .

Note:
While solving this problem, do all the steps properly to avoid confusion and mathematical errors. Do remember all the trigonometric formulas and mention them in the steps, wherever applicable, to get an accurate answer.