Question

What is `tan(pi + arcsin (2/3))`?

Answer 1

`(2sqrt(5))/5`

Explanation:

First thing to note is that every `color(red)tan` function has a period of `pi`

This means that `tan(pi+color(green)"angle")-=tan(color(green)"angle")`

`=>tan(pi+arcsin(2/3))=tan(arcsin(2/3))`

Now, let `theta=arcsin(2/3)`

So, now we are looking for `color(red)tan(theta)!`

We also have it that : `sin(theta)=2/3`

Next, we use the identity : `tan(theta)=sin(theta)/cos(theta)=sin(theta)/sqrt(1-sin^2(theta))`

And then we substitute the value for `sin(theta)`

`=>tan(theta)=(2/3)/sqrt(1-(2/3)^2)=2/3xx1/sqrt(1-4/9)=2/3xx1/sqrt((9-4)/9)=2/3xxsqrt(9/(9-4))=2/3xx3/sqrt(5)=2/sqrt(5)=(2sqrt(5))/5`