Question #c5fcd

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)#.

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.