Question

How do you evaluate $ \tan \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) $ ?

Answer 1

Hint: Here we will be equating our given trigonometric value to an unknown variable $ x $ and move $ \tan $ to the right hand side of the equation. The tan will become $ {\tan ^{ - 1}} $ and later we will be changing $ {\sin ^{ - 1}} $ into tan-1 for equating left hand side and right hand side.

Complete step-by-step answer:
Let $ \tan \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) $ be equal to $ x $ ,
 $ \tan \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) = x $
Now taking tan to right side, it will become $ {\tan ^{ - 1}} $ ,
 $ \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) = {\tan ^{ - 1}}x $ , let this be equation $ 1 $
Now solving left hand side using the identity $ {\sin ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {1 - {x^2}} }}} \right) $ we will have, here $ x = \dfrac{2}{3} $
 $ {\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{3}}}{{\sqrt {1 - {{(\dfrac{2}{3})}^2}} }}} \right) $ ,
Squaring $ \dfrac{2}{3} $ inside the root,
\[{\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{3}}}{{\sqrt {1 - (\dfrac{4}{9})} }}} \right)\]
Taking the L.C.M inside the root,
\[{\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{3}}}{{\sqrt {\dfrac{{9 - 4}}{9}} }}} \right)\]
\[{\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{3}}}{{\sqrt {\dfrac{5}{9}} }}} \right)\]
\[{\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{3}}}{{\dfrac{{\sqrt 5 }}{3}}}} \right)\]
\[{\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{2}{3} \times \dfrac{3}{{\sqrt 5 }}} \right)\]
\[{\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{2}{{\sqrt 5 }}} \right)\]
Now putting this value of \[{\sin ^{ - 1}}(\dfrac{2}{3})\]in equation $ 1 $ will give us,
 $ {\tan ^{ - 1}}(\dfrac{2}{{\sqrt 5 }}) = {\tan ^{ - 1}}x $
As $ {\tan ^{ - 1}} $ is at both sides, hence it will get cancel and we will get our value of x as $ \dfrac{2}{{\sqrt 5 }} $ and this is equal to $ \tan \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) $ .
So, $ \dfrac{2}{{\sqrt 5 }} $ is our desired answer.\[\]
So, the correct answer is “ $ \dfrac{2}{{\sqrt 5 }} $ ”.

Note: While moving the trigonometric terms from left hand side to right side or vice versa one should pay keen attention to the change in the trigonometric forms as here tan was changed into $ {\tan ^{ - 1}} $ .