Question

# If $\left( {{{\cos }^{ - 1}}x} \right) + \left( {{{\cos }^{ - 1}}y} \right) = 2\pi$, then find the value of $\left( {{{\sin }^{ - 1}}x} \right) + \left( {{{\sin }^{ - 1}}y} \right)$.${\text{A}}{\text{. }}\pi \\ {\text{B}}{\text{. }} - \pi \\ {\text{C}}{\text{. }}\dfrac{\pi }{2} \\$ ${\text{D}}{\text{.}}$ None of these

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Hint- Here, we will proceed by using the important inverse trigonometric identity which is $\left( {{{\sin }^{ - 1}}x} \right) + \left( {{{\cos }^{ - 1}}x} \right) = \dfrac{\pi }{2}$ where x is any value once in such a way that the given equation reduces to an equation from where the value of the required expression can be found.

Given, $\left( {{{\cos }^{ - 1}}x} \right) + \left( {{{\cos }^{ - 1}}y} \right) = 2\pi {\text{ }} \to {\text{(1)}}$
We have to find the value of the expression $\left( {{{\sin }^{ - 1}}x} \right) + \left( {{{\sin }^{ - 1}}y} \right)$
According to inverse trigonometric identities, we know that the sum of the inverse sine trigonometric function of any value with the inverse cosine trigonometric function of the same value will always be equal to $\dfrac{\pi }{2}$
For any value x, $\left( {{{\sin }^{ - 1}}x} \right) + \left( {{{\cos }^{ - 1}}x} \right) = \dfrac{\pi }{2}$
Taking $\left( {{{\sin }^{ - 1}}x} \right)$ from the LHS of the above equation to the RHS of the above equation, we get
$\left( {{{\cos }^{ - 1}}x} \right) = \dfrac{\pi }{2} - \left( {{{\sin }^{ - 1}}x} \right){\text{ }} \to {\text{(2)}}$
For any value y, $\left( {{{\sin }^{ - 1}}y} \right) + \left( {{{\cos }^{ - 1}}y} \right) = \dfrac{\pi }{2}$
Taking $\left( {{{\sin }^{ - 1}}y} \right)$ from the LHS of the above equation to the RHS of the above equation, we get
$\left( {{{\cos }^{ - 1}}y} \right) = \dfrac{\pi }{2} - \left( {{{\sin }^{ - 1}}y} \right){\text{ }} \to {\text{(3)}}$
By substituting the values of $\left( {{{\cos }^{ - 1}}x} \right)$ and $\left( {{{\cos }^{ - 1}}y} \right)$ from the equations (2) and (3) in the equations (1), we get
$\Rightarrow \dfrac{\pi }{2} - \left( {{{\sin }^{ - 1}}x} \right) + \dfrac{\pi }{2} - \left( {{{\sin }^{ - 1}}y} \right) = 2\pi \\ \Rightarrow \pi - \left( {{{\sin }^{ - 1}}x} \right) - \left( {{{\sin }^{ - 1}}y} \right) = 2\pi \\ \Rightarrow \left( {{{\sin }^{ - 1}}x} \right) + \left( {{{\sin }^{ - 1}}y} \right) = \pi - 2\pi \\ \Rightarrow \left( {{{\sin }^{ - 1}}x} \right) + \left( {{{\sin }^{ - 1}}y} \right) = - \pi \\$
Therefore, the value of the required expression $\left( {{{\sin }^{ - 1}}x} \right) + \left( {{{\sin }^{ - 1}}y} \right)$ is $- \pi$ radians.
Hence, option B is correct.

Note- Apart from the identity $\left( {{{\sin }^{ - 1}}x} \right) + \left( {{{\cos }^{ - 1}}x} \right) = \dfrac{\pi }{2}$, there are two other inverse trigonometric identities of the same form which are $\left( {{{\tan }^{ - 1}}x} \right) + \left( {{{\cot }^{ - 1}}x} \right) = \dfrac{\pi }{2}$ and $\left( {{{\sec }^{ - 1}}x} \right) + \left( {{{\operatorname{cosec} }^{ - 1}}x} \right) = \dfrac{\pi }{2}$ for any value x. These identities can be given to convert any equation having inverse tangent and inverse secant trigonometric functions into inverse cotangent and inverse cosecant trigonometric functions respectively and its vice versa.