Question

The solution of the equation ${\sin ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) - {\sin ^{ - 1}}\left( {\sqrt {\dfrac{3}{x}} } \right) - \dfrac{\pi }{6} = 0$ is given by $x = $
A) $1$
B) $2$
C) $3$
D) $4$

Answer 1

Hint: In this question we have been given an equation ${\sin ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) - {\sin ^{ - 1}}\left( {\sqrt {\dfrac{3}{x}} } \right) - \dfrac{\pi }{6} = 0$. We can see that there are trigonometric ratios in these types of equations, so we will use the trigonometric identities to solve this question.
We will try to eliminate the sine from the equation, so we will first use the values. We know that the value of $\tan \dfrac{\pi }{4}$ can be also be written as $\tan \dfrac{{180}}{4} = \tan {45^ \circ }$
And we know that the value of $\tan {45^ \circ } = 1$ . So we can say that $\tan \dfrac{\pi }{4} = 1$ .

Complete step by step solution:
Here we have ${\sin ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) - {\sin ^{ - 1}}\left( {\sqrt {\dfrac{3}{x}} } \right) - \dfrac{\pi }{6} = 0$
We know that the value of
$\tan \dfrac{\pi }{4} = 1$
So we can write the equation as
${\sin ^{ - 1}}\left( 1 \right) - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) - \dfrac{\pi }{6} = 0$ .
We should keep in mind that ${\sin ^{ - 1}}(1)$ is equal to the angle whose sine value is $1$ .
And we know that the value of $\sin {90^ \circ } = 1$
So we can say that the inverse of ${\sin ^{ - 1}}(1)$ is ${90^ \circ }$.
We can also write $90$ as $\dfrac{\pi }{2}$
Now we can write the expression as
 $\dfrac{\pi }{2} - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) - \dfrac{\pi }{6} = 0$ .
We will take the similar terms to the other sides of the equation and we have :
$ - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) = \dfrac{\pi }{6} - \dfrac{\pi }{2}$ .
On simplifying we have :
$ - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) = \dfrac{{\pi - 3\pi }}{6} \Rightarrow - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) = - \dfrac{{2\pi }}{6}$ .
After eliminating the negative sign, the equation can also be written as :
${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right) = \dfrac{\pi }{3}$
By taking $\sin $on both the sides of the equation we have:
$\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right)} \right) = \sin \dfrac{\pi }{3}$
We know the trigonometric identity that
$\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta $
So we can write
$\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{{\sqrt x }}} \right)} \right) = \dfrac{{\sqrt 3 }}{x}$.
And the value of $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
By putting the value back in the equation we have :
$\dfrac{{\sqrt 3 }}{{\sqrt x }} = \dfrac{{\sqrt 3 }}{2}$
By cross multiplication we have;
$\sqrt x = \dfrac{{2\sqrt 3 }}{{\sqrt 3 }} \Rightarrow \sqrt x = 2$
We can square both the sides to eliminate the under root sign i.e.
${\left( {\sqrt x } \right)^2} = {2^2}$
So it gives us value $x = 4$
Hence the correct option is (D) $4$.

Note:
We should note that
$\pi = {180^ \circ }$ .
So in the above solution, we can write $\dfrac{\pi }{3}$ as
 $\dfrac{{180}}{3} = 60$
It gives us
$\sin {60^ \circ }$
And we know the trigonometric value that
$\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$ .
So it gives us the value of
$\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ .