Question

If${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \dfrac{{2\pi }}{3}$,${\cos ^{ - 1}}x - {\cos ^{ - 1}}y = - \dfrac{\pi }{3}$, then the number of values of$\left( {x,y} \right)$is
A. $2$
B. $4$
C. $0$
D. None of these

Answer 1

Hint: 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$\left( {x,y} \right)$. 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 $x$ and$y$.
Formula to be used:
The trigonometric identity that we need to apply in this problem is as follows.
${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$

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

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