Question

Question #c5fcd

Answer 1

`x/sqrt(1-x^2)`

Explanation:

Draw a right triangle with an angle `theta`, the hypotenuse `1`, and the opposite side `x`. Notice that `arcsin(theta)=x/1=x`.

Now, we want to find the tangent of the angle `theta`. For this, we need to find the adjacent side. We can do this by using the Pythagorean Theorem: `sqrt(1-x^2)`. The tangent of angle `\theta` is the opposite side divided by the adjacent side, or `x/sqrt(1-x^2)`.

Answer 2

`x/sqrt(1-x^2)`

Explanation:

Here is another solution other than the one I posted above. We will use the identities `1+tan^2(theta)=sec^2(theta)` and `sin^2(theta)+cos^2(theta)=1`.

From the first identity, we see that `tan(theta)=sqrt(sec^2(theta)-1)`. We can rewrite `tan(sin^-1(x))` as `sqrt(sec^2(sin^-1(x))-1)`, which is equal to `sqrt(1/cos^2(sin^-1(x))-1)`.

Now, by our second identity, `cos^2(theta)=1-sin^2(theta)`. We can then rewrite `sqrt(1/cos^2(sin^-1(x))-1)` as `sqrt(1/(1-sin^2(sin^-1(x)))-1)`. This can be furthered simplified to `sqrt(1/(1-x^2)-1)=sqrt((1-(1-x^2))/(1-x^2))=sqrt(x^2/(1-x^2))=x/sqrt(1-x^2)`.

If you want the combination of trigonometric functions of inverse trigonometric functions, Wikipedia has a nice table here.