Question

If ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = {\tan ^{ - 1}}x$. Then find the value of $x$.
A. $\sqrt 3 $
B. $\dfrac{1}{{\sqrt 3 }}$
C. $\dfrac{1}{{\sqrt 2 }}$
D. None of these

Answer 1

Hint: First we will simplify the left side of the given equation by converting $\dfrac{1}{2}$ into $\sin \theta $ form. Then apply the inverse formula to find the value of x.

Formula Used:
$\sin \dfrac{\pi }{6} = \dfrac{1}{2}$
${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta $
${\tan ^{ - 1}}x = a \Rightarrow x = \tan a$

Complete step by step solution:
Given trigonometry equation is
${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = {\tan ^{ - 1}}x$
Putting the $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$
${\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{6}} \right) = {\tan ^{ - 1}}x$
Now we will apply the formula ${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta $ on the left side of the expression
$\dfrac{\pi }{6} = {\tan ^{ - 1}}x$
Apply the formula ${\tan ^{ - 1}}x = a \Rightarrow x = \tan a$
$ \Rightarrow x = \tan \dfrac{\pi }{6}$
Putting the value of $\tan \dfrac{\pi }{6}$.
$ \Rightarrow x = \dfrac{1}{{\sqrt 3 }}$

Option ‘B’ is correct

Note: To solve this question, students must be well versed in $\sin\theta$ and $\tan\theta$ table and inverse formula. In this question, after substituting $\dfrac{1}{2}=\sin \dfrac{\pi}{6}$, you need to apply the arcsin of sin formula. Remember one thing: don't write $\tan^{-1}=\dfrac{1}{\tan}$, because $\tan^{-1}$ is a function.