Question

How do you evaluate $\tan \left( \arcsin \left( \dfrac{1}{3} \right) \right)$?

Answer 1

Hint: First express the expression in terms of sine and cosine form using the formula $\tan x=\dfrac{\sin x}{\cos x}$. Then simplify the numerator and the denominator by taking them separately. For numerator use the formula $\sin \left( \arcsin \left( x \right) \right)=x$and for the denominator use $\cos x=\sqrt{1-{{\sin }^{2}}x}$ and ${{\sin }^{2}}\left( \arcsin \left( x \right) \right)={{x}^{2}}$. After getting simplified values, bring the numerator and the denominator together and do the necessary calculation to obtain the required solution.

Complete step by step solution:
 As we know, $\tan x=\dfrac{\sin x}{\cos x}$
Now, considering our question $\tan \left( \arcsin \left( \dfrac{1}{3} \right) \right)$
It can be written as $\tan \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{\sin \left( \arcsin \left( \dfrac{1}{3} \right) \right)}{\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)}$
From here, we have to simplify the numerator part and the denominator part separately.
For the numerator part:
As we know, $\sin \left( \arcsin \left( x \right) \right)=x$
So, $\sin \left( \arcsin \left( \dfrac{1}{3} \right) \right)$ can be written as $\sin \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{1}{3}$
For the denominator part:
As we know
 $\begin{align}
  & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & \Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x \\
 & \Rightarrow \cos x=\sqrt{1-{{\sin }^{2}}x} \\
\end{align}$
So, $\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)$ can be written as $\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\sqrt{1-{{\sin }^{2}}\left( \arcsin \left( \dfrac{1}{3} \right) \right)}$
Again, we know ${{\sin }^{2}}\left( \arcsin \left( x \right) \right)={{\left( x \right)}^{2}}={{x}^{2}}$
So, ${{\sin }^{2}}\left( \arcsin \left( \dfrac{1}{3} \right) \right)={{\left( \dfrac{1}{3} \right)}^{2}}=\dfrac{1}{9}$
Putting the value of ${{\sin }^{2}}\left( \arcsin \left( \dfrac{1}{3} \right) \right)$in $\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\sqrt{1-{{\sin }^{2}}\left( \arcsin \left( \dfrac{1}{3} \right) \right)}$, we get
$\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{9-1}{9}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt{2}}{3}$
Bringing the numerator and the denominator together, now our expression becomes
$\tan \left( \arcsin \left( \dfrac{1}{3} \right) \right)=\dfrac{\sin \left( \arcsin \left( \dfrac{1}{3} \right) \right)}{\cos \left( \arcsin \left( \dfrac{1}{3} \right) \right)}=\dfrac{\dfrac{1}{3}}{\dfrac{2\sqrt{2}}{3}}=\dfrac{1}{3}\times \dfrac{3}{2\sqrt{2}}=\dfrac{1}{2\sqrt{2}}$
This is the required solution of the given question.

Note: We know, ${{\sin }^{2}}x$ can be written as ${{\left( \sin x \right)}^{2}}$. Similarly we can write ${{\sin }^{2}}\left( \arcsin \left( x \right) \right)$ as ${{\left[ \sin \left( \arcsin \left( x \right) \right) \right]}^{2}}$. Again as we know, the value of $\sin \left( \arcsin \left( x \right) \right)=x$ so, the value of ${{\left[ \sin \left( \arcsin \left( x \right) \right) \right]}^{2}}={{\left( x \right)}^{2}}={{x}^{2}}$, which is used in this question. The above question should be solved by taking the numerator and the denominator separately to avoid errors and complex calculations.