Question

What is the value of \[\tan \left( {{{\sin }^{ - 1}}x} \right)\] ?

Answer 1

Hint: Here we will use the functions of trigonometry as well as the substitution to solve the given function. We will substitute \[{\sin ^{ - 1}}x = \theta \] . from this we will find the value of x. then we know that tan function is the ratio of sin and cos. So we will find the value of cos also. Then their ratio will be the required answer.

Complete step by step solution:
We are given that \[\tan \left( {{{\sin }^{ - 1}}x} \right)\]
Let \[{\sin ^{ - 1}}x = \theta \]
Then applying sin function on both sides we get,
\[x = \sin \theta \]
Now we know that,
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
So we need to find the value of the cos function.
\[\cos \theta = \sqrt {1 - {{\sin }^2}\theta } \]
But, \[x = \sin \theta \]
So,
\[\cos \theta = \sqrt {1 - {x^2}} \]
Now taking the ratio as,
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Now substituting the values,
\[\tan \theta = \dfrac{x}{{\sqrt {1 - {x^2}} }}\]
And now substituting the value of theta,
\[\tan \left( {{{\sin }^{ - 1}}x} \right) = \dfrac{x}{{\sqrt {1 - {x^2}} }}\]
Therefore, the value of \[\tan \left( {{{\sin }^{ - 1}}x} \right) = \dfrac{x}{{\sqrt {1 - {x^2}} }}\].

Note:
Note that we are not asked to find the value for any particular angle. If so given we have expressed the values in theta form also. We need to just put the angle there and find the correct value. Also note that tan function is the ratio of sin and cos, so we can find the ratio using a right angle triangle whose hypotenuse is 1 and one of its sides is x, then we can find the third side and then any trigonometric function.