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

Ifsin⁡−1x+sin⁡−1y=2π3{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3},cos⁡−1x−cos⁡−1y=−π3{\cos ^{ - 1}}x - {\cos ^{ - 1}}y = - \dfrac{\pi }{3}, then the number of values of(x,y)\left( {x,y} \right)is
A. 22
B. 44
C. 00
D. None of these

Explanation

Solution

We need to analyze the given information first, so that we are able to solve the given problem. Here we are asked to calculate the number of values of(x,y)\left( {x,y} \right). The number of values refers to the solution. That is we need to calculate the solution for the given equations. To find the solution, the first step we need to follow is to solve the given equations. Here, we are going to solve the given equation to obtain the values of xx andyy.
Formula to be used:
The trigonometric identity that we need to apply in this problem is as follows.
sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}

Complete step by step answer:
Let sin⁡−1x+sin⁡−1y=2π3{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3} …..(1)\left( 1 \right) andcos⁡−1x−cos⁡−1y=−π3{\cos ^{ - 1}}x - {\cos ^{ - 1}}y = - \dfrac{\pi }{3} ….(2)\left( 2 \right)
We shall add the equation (1)\left( 1 \right) and the equation (2)\left( 2 \right)
That is sin⁡−1x+sin⁡−1y+cos⁡−1x−cos⁡−1y=2π3−π3{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\cos ^{ - 1}}x - {\cos ^{ - 1}}y = \dfrac{{2\pi }}{3} - \dfrac{\pi }{3}
(sin⁡−1x+cos⁡−1x)+(sin⁡−1y−cos⁡−y)=π3\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right) + \left( {{{\sin }^{ - 1}}y - {{\cos }^ - }y} \right) = \dfrac{\pi }{3}
π2+(sin⁡−1y−cos⁡−1y)=π3\dfrac{\pi }{2} + \left( {{{\sin }^{ - 1}}y - {{\cos }^{ - 1}}y} \right) = \dfrac{\pi }{3} (Here we applied the trigonometric identitysin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2} )
Hence, we get sin⁡−1y−cos⁡−1y=−π6{\sin ^{ - 1}}y - {\cos ^{ - 1}}y = - \dfrac{\pi }{6} …….(3)\left( 3 \right)
We all know the following trigonometric identity sin⁡−1y+cos⁡−1y=π2{\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \dfrac{\pi }{2} ….(4)\left( 4 \right)
Now we need to add the equation (3)\left( 3 \right) and the equation (4)\left( 4 \right)
sin⁡−1y−cos⁡−1y+sin⁡−1y+cos⁡−1y=−π6+π2{\sin ^{ - 1}}y - {\cos ^{ - 1}}y + {\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \dfrac{{ - \pi }}{6} + \dfrac{\pi }{2}
⇒2sin⁡−1y=−π+3π6\Rightarrow 2{\sin ^{ - 1}}y = \dfrac{{ - \pi + 3\pi }}{6}
⇒2sin⁡−1y=2π6\Rightarrow 2{\sin ^{ - 1}}y = \dfrac{{2\pi }}{6}
⇒sin⁡−1y=π3×12\Rightarrow {\sin ^{ - 1}}y = \dfrac{\pi }{3} \times \dfrac{1}{2}
=π6= \dfrac{\pi }{6}
sin⁡−1y=π6{\sin ^{ - 1}}y = \dfrac{\pi }{6}
⇒y=sin⁡π6\Rightarrow y = \sin \dfrac{\pi }{6}
Thus, we gety=12y = \dfrac{1}{2}
Now we need to substitute sin⁡−1y=π6{\sin ^{ - 1}}y = \dfrac{\pi }{6} in the equation(1)\left( 1 \right)
sin⁡−1x+sin⁡−1y=2π3{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}
⇒sin⁡−1x+π6=2π3\Rightarrow {\sin ^{ - 1}}x + \dfrac{\pi }{6} = \dfrac{{2\pi }}{3}
⇒sin⁡−1x=2π3−π6\Rightarrow {\sin ^{ - 1}}x = \dfrac{{2\pi }}{3} - \dfrac{\pi }{6}
=4π−π6= \dfrac{{4\pi - \pi }}{6}
=3π6= \dfrac{{3\pi }}{6}
sin⁡−1x=π2{\sin ^{ - 1}}x = \dfrac{\pi }{2}
x=sin⁡π2x = \sin \dfrac{\pi }{2}
⇒x=1\Rightarrow x = 1
Therefore, the obtained solution is (x,y)=(1,12)\left( {x,y} \right) = \left( {1,\dfrac{1}{2}} \right),So the number of values of (x,y)\left( {x,y} \right) is 1.
So, the correct answer is “Option D”.

Note: We have obtained the solution (x,y)=(1,12)\left( {x,y} \right) = \left( {1,\dfrac{1}{2}} \right).we can also verify whether the obtained solution is correct or not. We shall substitutex=1x = 1 ,y=12y = \dfrac{1}{2} in the equation (2)\left( 2 \right)
That is cos⁡−1x−cos⁡−1y=−π3{\cos ^{ - 1}}x - {\cos ^{ - 1}}y = - \dfrac{\pi }{3}
⇒cos⁡−1(1)−cos⁡−1(12)=−π3\Rightarrow {\cos ^{ - 1}}\left( 1 \right) - {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = - \dfrac{\pi }{3}
⇒0−π3=−π3\Rightarrow 0 - \dfrac{\pi }{3} = - \dfrac{\pi }{3}
Hence we get −π3 - \dfrac{\pi }{3} which is on the right-hand side of the equation.
Therefore, the obtained solution (x,y)=(1,12)\left( {x,y} \right) = \left( {1,\dfrac{1}{2}} \right)is verified.