So, we want to express the following:
![\sec (\sin ^(-1)(\frac{x}{\sqrt[]{x^2+81}}))](https://img.qammunity.org/2023/formulas/mathematics/college/qcx4m9ecfzh2ebbs6du6u4jsvuooc18kem.png)
As an algebraic expression.
If:
![\begin{gathered} \sin ^(-1)(\frac{x}{\sqrt[]{x^2+81}})=\theta \\ \text{Then,} \\ \sin (\theta)=\frac{x}{\sqrt[]{x^2+81}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hs55anirp95fm724sn7ge8i6yfdjp4zfw2.png)
We could draw the following triangle:
Remember that the secant function relations the hypotenuse of the triangle and the adjacent side of the triangle. So first, we should find the adjacent side using the pythagorean theorem:
![\begin{gathered} a^2=(\sqrt[]{x^2+81})^2-x^2 \\ a^2=x^2+81-x^2 \\ a^2=81\to a=9 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vtuy7a845jd45ekry53lpsw3mkl24c78kq.png)
Therefore, the adjacent side is 81. And, the value of:
![\sec (\sin ^(-1)(\frac{x}{\sqrt[]{x^2+81}}))](https://img.qammunity.org/2023/formulas/mathematics/college/qcx4m9ecfzh2ebbs6du6u4jsvuooc18kem.png)
Is:
![\sec (\sin ^(-1)(\frac{x}{\sqrt[]{x^2+81}}))=\frac{\sqrt[]{x^2+81}}{9}](https://img.qammunity.org/2023/formulas/mathematics/college/qnmgqjm5cc80ldzkxgsltou07c9qhxuryu.png)