Question

If \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}\], then \[{\cos ^{ - 1}}x + {\cos ^{ - 1}}y\] is equal to
(a) \[\dfrac{{2\pi }}{3}\]
(b) \[\dfrac{\pi }{3}\]
(c) \[\dfrac{\pi }{6}\]
(d) \[\pi \]

Answer 1

Hint:
Here, we need to find the value of the expression \[{\cos ^{ - 1}}x + {\cos ^{ - 1}}y\]. We will use the formula for sum of the sine inverse of an angle and the cosine inverse of an angle. Then, we will rewrite the given equation. Finally, we will simplify the equation to find the value of the expression \[{\cos ^{ - 1}}x + {\cos ^{ - 1}}y\].
Formula Used: The sum of the sine inverse of an angle \[\theta \] and the cosine inverse of the angle \[\theta \] is given by \[{\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}\].

Complete step by step solution:
We will use the formula for sum of the sine inverse of an angle and the cosine inverse of an angle.
The sum of the sine inverse of an angle \[\theta \] and the cosine inverse of the angle \[\theta \] is given by \[{\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}\].
Substituting \[\theta = x\] in the formula \[{\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}\], we get
\[ \Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\]
Subtracting \[{\cos ^{ - 1}}x\] from both sides of the equation, we get
\[ \Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x - {\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\cos ^{ - 1}}x\]
Thus, we get
\[ \Rightarrow {\sin ^{ - 1}}x = \dfrac{\pi }{2} - {\cos ^{ - 1}}x\]
Substituting \[\theta = y\] in the formula \[{\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}\], we get
\[ \Rightarrow {\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \dfrac{\pi }{2}\]
Subtracting \[{\cos ^{ - 1}}y\] from both sides of the equation, we get
\[ \Rightarrow {\sin ^{ - 1}}y + {\cos ^{ - 1}}y - {\cos ^{ - 1}}y = \dfrac{\pi }{2} - {\cos ^{ - 1}}y\]
Thus, we get
\[ \Rightarrow {\sin ^{ - 1}}y = \dfrac{\pi }{2} - {\cos ^{ - 1}}y\]
Now, we will find the value of the expression \[{\cos ^{ - 1}}x + {\cos ^{ - 1}}y\].
Substituting \[{\sin ^{ - 1}}x = \dfrac{\pi }{2} - {\cos ^{ - 1}}x\] and \[{\sin ^{ - 1}}y = \dfrac{\pi }{2} - {\cos ^{ - 1}}y\] in the equation \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}\], we get
\[ \Rightarrow \dfrac{\pi }{2} - {\cos ^{ - 1}}x + \dfrac{\pi }{2} - {\cos ^{ - 1}}y = \dfrac{{2\pi }}{3}\]
Adding the terms of the expression, we get
\[ \Rightarrow \dfrac{{\pi + \pi }}{2} - {\cos ^{ - 1}}x - {\cos ^{ - 1}}y = \dfrac{{2\pi }}{3}\]
Thus, we get
\[\begin{array}{l} \Rightarrow \dfrac{{2\pi }}{2} - {\cos ^{ - 1}}x - {\cos ^{ - 1}}y = \dfrac{{2\pi }}{3}\\ \Rightarrow \pi - {\cos ^{ - 1}}x - {\cos ^{ - 1}}y = \dfrac{{2\pi }}{3}\end{array}\]
Rewriting the expression, we get
\[ \Rightarrow \pi - \dfrac{{2\pi }}{3} = {\cos ^{ - 1}}x + {\cos ^{ - 1}}y\]
We can rewrite the equation as
\[ \Rightarrow \dfrac{{3\pi }}{3} - \dfrac{{2\pi }}{3} = {\cos ^{ - 1}}x + {\cos ^{ - 1}}y\]
Subtracting the terms of the expression, we get
\[ \Rightarrow \dfrac{{3\pi - 2\pi }}{3} = {\cos ^{ - 1}}x + {\cos ^{ - 1}}y\]
Thus, we get
\[ \Rightarrow \dfrac{\pi }{3} = {\cos ^{ - 1}}x + {\cos ^{ - 1}}y\]
Therefore, we get the value of the expression \[{\cos ^{ - 1}}x + {\cos ^{ - 1}}y\] as \[\dfrac{\pi }{3}\].

Thus, the correct option is option (b).

Note:
We can rewrite the equations \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\] and \[{\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \dfrac{\pi }{2}\] as \[{\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x\] and \[{\cos ^{ - 1}}y = \dfrac{\pi }{2} - {\sin ^{ - 1}}y\] respectively.
Adding the two equations, we get
\[ \Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{\pi }{2} - {\sin ^{ - 1}}x + \dfrac{\pi }{2} - {\sin ^{ - 1}}y\]
Adding the terms, we get
\[\begin{array}{l} \Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{{\pi + \pi }}{2} - {\sin ^{ - 1}}x - {\sin ^{ - 1}}y\\ \Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{{2\pi }}{2} - {\sin ^{ - 1}}x - {\sin ^{ - 1}}y\\ \Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \pi - {\sin ^{ - 1}}x - {\sin ^{ - 1}}y\end{array}\]
Factoring out \[ - 1\] from the terms of the expression, we get
\[ \Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \pi - 1\left( {{{\sin }^{ - 1}}x + {{\sin }^{ - 1}}y} \right)\]
It is given that \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}\].
Substituting \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}\] in the equation, we get
\[\begin{array}{l} \Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \pi - 1\left( {\dfrac{{2\pi }}{3}} \right)\\ \Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \pi - \dfrac{{2\pi }}{3}\end{array}\]
The L.C.M. of 1 and 3 is 3.
Rewriting the terms with a denominator of 3, we get
\[ \Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{{3\pi }}{3} - \dfrac{{2\pi }}{3}\]
Subtracting the terms, we get
\[ \Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{\pi }{3}\]
Therefore, we get the value of the expression \[{\cos ^{ - 1}}x + {\cos ^{ - 1}}y\] as \[\dfrac{\pi }{3}\].
Thus, the correct option is option (b).