Answer:
![\csc(x)=5/3](https://img.qammunity.org/2021/formulas/mathematics/high-school/ar6kainassvvz1lw6huinn93a9n5ca9qfg.png)
Explanation:
So we know that:
![\tan(x)=3/4](https://img.qammunity.org/2021/formulas/mathematics/high-school/9fpnej113941hw4pxovwz2ii0vpe3lek9w.png)
Recall that tangent represents the side opposite over the side adjacent.
This means that the opposite side is 3 while the adjacent side is 4.
Therefore, by using the Pythagorean Theorem, we can solve for the third side, which is the hypotenuse:
![a^2+b^2=c^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/96dopf217hvzc3zhswffnjr8l5f26vmjhb.png)
Substitute a for 3 and b for 4. Thus:
![3^2+4^2=c^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/dzx28y70spx5fsoqs4mnvk9uza465gl4jl.png)
Square and add:
![9+16=c^2\\25=c^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/j0xzaws5kt4y37uqtwfzpb6bglk506bf3n.png)
Square root:
![c=5](https://img.qammunity.org/2021/formulas/mathematics/middle-school/b024tcrunfq9n180wl6u91mg6zpb9qjcul.png)
Now, recall that cosecant is the reciprocal of sine. So, find sine first.
Sine is opposite over hypotenuse. From tangent, the opposite is 3 and the hypotenuse as we now know is 5. Thus:
![\sin(x)=3/5](https://img.qammunity.org/2021/formulas/mathematics/high-school/8lkk3lbpz1frg2e9xtp8runan37ugqwn30.png)
And cosecant is the reciprocal of that. Thus:
![\csc(x)=5/3](https://img.qammunity.org/2021/formulas/mathematics/high-school/ar6kainassvvz1lw6huinn93a9n5ca9qfg.png)
And that's our answer :)