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

How do you evaluate tan⁡(sin⁡−1(23))\tan \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) ?

Explanation

Solution

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

Complete step-by-step answer :
Let tan⁡(sin⁡−1(23))\tan \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) be equal to xx ,
tan⁡(sin⁡−1(23))=x\tan \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) = x
Now taking tan to right side, it will become tan⁡−1{\tan ^{ - 1}} ,
(sin⁡−1(23))=tan⁡−1x\left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) = {\tan ^{ - 1}}x , let this be equation 11
Now solving left hand side using the identity sin⁡−1x=tan⁡−1(x1−x2){\sin ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{x}{{\sqrt {1 - {x^2}} }}} \right) we will have, here x=23x = \dfrac{2}{3}
sin⁡−1(23)=tan⁡−1(231−(23)2){\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{3}}}{{\sqrt {1 - {{(\dfrac{2}{3})}^2}} }}} \right) ,
Squaring 23\dfrac{2}{3} inside the root,
sin⁡−1(23)=tan⁡−1(231−(49)){\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(23)=tan⁡−1(239−49){\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{3}}}{{\sqrt {\dfrac{{9 - 4}}{9}} }}} \right)
sin⁡−1(23)=tan⁡−1(2359){\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{3}}}{{\sqrt {\dfrac{5}{9}} }}} \right)
sin⁡−1(23)=tan⁡−1(2353){\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{3}}}{{\dfrac{{\sqrt 5 }}{3}}}} \right)
sin⁡−1(23)=tan⁡−1(23×35){\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{2}{3} \times \dfrac{3}{{\sqrt 5 }}} \right)
sin⁡−1(23)=tan⁡−1(25){\sin ^{ - 1}}(\dfrac{2}{3}) = {\tan ^{ - 1}}\left( {\dfrac{2}{{\sqrt 5 }}} \right)
Now putting this value of sin⁡−1(23){\sin ^{ - 1}}(\dfrac{2}{3})in equation 11 will give us,
tan⁡−1(25)=tan⁡−1x{\tan ^{ - 1}}(\dfrac{2}{{\sqrt 5 }}) = {\tan ^{ - 1}}x
As tan⁡−1{\tan ^{ - 1}} is at both sides, hence it will get cancel and we will get our value of x as 25\dfrac{2}{{\sqrt 5 }} and this is equal to tan⁡(sin⁡−1(23))\tan \left( {{{\sin }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) .
So, 25\dfrac{2}{{\sqrt 5 }} is our desired answer.$$$$
So, the correct answer is “ 25\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{\tan ^{ - 1}} .