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Question: If \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}\], then \[{\cos ^{ - 1}}x + {\cos ^{ - 1...

If sin⁡−1x+sin⁡−1y=2π3{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}, then cos⁡−1x+cos⁡−1y{\cos ^{ - 1}}x + {\cos ^{ - 1}}y is equal to
(a) 2π3\dfrac{{2\pi }}{3}
(b) π3\dfrac{\pi }{3}
(c) π6\dfrac{\pi }{6}
(d) π\pi

Explanation

Solution

Here, we need to find the value of the expression cos⁡−1x+cos⁡−1y{\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⁡−1x+cos⁡−1y{\cos ^{ - 1}}x + {\cos ^{ - 1}}y.
Formula Used: The sum of the sine inverse of an angle θ\theta and the cosine inverse of the angle θ\theta is given by sin⁡−1θ+cos⁡−1θ=π2{\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 θ\theta and the cosine inverse of the angle θ\theta is given by sin⁡−1θ+cos⁡−1θ=π2{\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}.
Substituting θ=x\theta = x in the formula sin⁡−1θ+cos⁡−1θ=π2{\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}, we get
⇒sin⁡−1x+cos⁡−1x=π2\Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}
Subtracting cos⁡−1x{\cos ^{ - 1}}x from both sides of the equation, we get
⇒sin⁡−1x+cos⁡−1x−cos⁡−1x=π2−cos⁡−1x\Rightarrow {\sin ^{ - 1}}x + {\cos ^{ - 1}}x - {\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\cos ^{ - 1}}x
Thus, we get
⇒sin⁡−1x=π2−cos⁡−1x\Rightarrow {\sin ^{ - 1}}x = \dfrac{\pi }{2} - {\cos ^{ - 1}}x
Substituting θ=y\theta = y in the formula sin⁡−1θ+cos⁡−1θ=π2{\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}, we get
⇒sin⁡−1y+cos⁡−1y=π2\Rightarrow {\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \dfrac{\pi }{2}
Subtracting cos⁡−1y{\cos ^{ - 1}}y from both sides of the equation, we get
⇒sin⁡−1y+cos⁡−1y−cos⁡−1y=π2−cos⁡−1y\Rightarrow {\sin ^{ - 1}}y + {\cos ^{ - 1}}y - {\cos ^{ - 1}}y = \dfrac{\pi }{2} - {\cos ^{ - 1}}y
Thus, we get
⇒sin⁡−1y=π2−cos⁡−1y\Rightarrow {\sin ^{ - 1}}y = \dfrac{\pi }{2} - {\cos ^{ - 1}}y
Now, we will find the value of the expression cos⁡−1x+cos⁡−1y{\cos ^{ - 1}}x + {\cos ^{ - 1}}y.
Substituting sin⁡−1x=π2−cos⁡−1x{\sin ^{ - 1}}x = \dfrac{\pi }{2} - {\cos ^{ - 1}}x and sin⁡−1y=π2−cos⁡−1y{\sin ^{ - 1}}y = \dfrac{\pi }{2} - {\cos ^{ - 1}}y in the equation sin⁡−1x+sin⁡−1y=2π3{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}, we get
⇒π2−cos⁡−1x+π2−cos⁡−1y=2π3\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
⇒π+π2−cos⁡−1x−cos⁡−1y=2π3\Rightarrow \dfrac{{\pi + \pi }}{2} - {\cos ^{ - 1}}x - {\cos ^{ - 1}}y = \dfrac{{2\pi }}{3}
Thus, we get
⇒2π2−cos⁡−1x−cos⁡−1y=2π3 ⇒π−cos⁡−1x−cos⁡−1y=2π3\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
⇒π−2π3=cos⁡−1x+cos⁡−1y\Rightarrow \pi - \dfrac{{2\pi }}{3} = {\cos ^{ - 1}}x + {\cos ^{ - 1}}y
We can rewrite the equation as
⇒3π3−2π3=cos⁡−1x+cos⁡−1y\Rightarrow \dfrac{{3\pi }}{3} - \dfrac{{2\pi }}{3} = {\cos ^{ - 1}}x + {\cos ^{ - 1}}y
Subtracting the terms of the expression, we get
⇒3π−2π3=cos⁡−1x+cos⁡−1y\Rightarrow \dfrac{{3\pi - 2\pi }}{3} = {\cos ^{ - 1}}x + {\cos ^{ - 1}}y
Thus, we get
⇒π3=cos⁡−1x+cos⁡−1y\Rightarrow \dfrac{\pi }{3} = {\cos ^{ - 1}}x + {\cos ^{ - 1}}y
Therefore, we get the value of the expression cos⁡−1x+cos⁡−1y{\cos ^{ - 1}}x + {\cos ^{ - 1}}y as π3\dfrac{\pi }{3}.

Thus, the correct option is option (b).

Note:
We can rewrite the equations sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2} and sin⁡−1y+cos⁡−1y=π2{\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \dfrac{\pi }{2} as cos⁡−1x=π2−sin⁡−1x{\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x and cos⁡−1y=π2−sin⁡−1y{\cos ^{ - 1}}y = \dfrac{\pi }{2} - {\sin ^{ - 1}}y respectively.
Adding the two equations, we get
⇒cos⁡−1x+cos⁡−1y=π2−sin⁡−1x+π2−sin⁡−1y\Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{\pi }{2} - {\sin ^{ - 1}}x + \dfrac{\pi }{2} - {\sin ^{ - 1}}y
Adding the terms, we get
⇒cos⁡−1x+cos⁡−1y=π+π2−sin⁡−1x−sin⁡−1y ⇒cos⁡−1x+cos⁡−1y=2π2−sin⁡−1x−sin⁡−1y ⇒cos⁡−1x+cos⁡−1y=π−sin⁡−1x−sin⁡−1y\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 - 1 from the terms of the expression, we get
⇒cos⁡−1x+cos⁡−1y=π−1(sin⁡−1x+sin⁡−1y)\Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \pi - 1\left( {{{\sin }^{ - 1}}x + {{\sin }^{ - 1}}y} \right)
It is given that sin⁡−1x+sin⁡−1y=2π3{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}.
Substituting sin⁡−1x+sin⁡−1y=2π3{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3} in the equation, we get
⇒cos⁡−1x+cos⁡−1y=π−1(2π3) ⇒cos⁡−1x+cos⁡−1y=π−2π3\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
⇒cos⁡−1x+cos⁡−1y=3π3−2π3\Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{{3\pi }}{3} - \dfrac{{2\pi }}{3}
Subtracting the terms, we get
⇒cos⁡−1x+cos⁡−1y=π3\Rightarrow {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \dfrac{\pi }{3}
Therefore, we get the value of the expression cos⁡−1x+cos⁡−1y{\cos ^{ - 1}}x + {\cos ^{ - 1}}y as π3\dfrac{\pi }{3}.
Thus, the correct option is option (b).