Answer:
1
Explanation:
First, convert all the secants and cosecants to cosine and sine, respectively. Recall that
and
.
Thus:
![(sec(x))/(cos(x)) -(sin(x))/(csc(x)cos^2(x))](https://img.qammunity.org/2021/formulas/mathematics/middle-school/glccvwykuajf52zm47fbw7lqrzctwhid34.png)
![=((1)/(cos(x)) )/(cos(x)) -(sin(x))/((1)/(sin(x))cos^2(x) )](https://img.qammunity.org/2021/formulas/mathematics/middle-school/fndtmwxauyer2osg8nob7ly4rftnstbkyu.png)
Let's do the first part first: (Recall how to divide fractions)
![((1)/(cos(x)) )/(cos(x))=(1)/(cos(x)) \cdot (1)/(cos(x))=(1)/(cos^2(x))](https://img.qammunity.org/2021/formulas/mathematics/middle-school/5ifvv7xvdghzq71mdwnvkcelx5crrnsyrs.png)
For the second term:
![(sin(x))/((cos^2(x))/(sin(x)) ) =(sin(x))/(1) \cdot(sin(x))/(cos^2(x))=(sin^2(x))/(cos^2(x))](https://img.qammunity.org/2021/formulas/mathematics/middle-school/1o4l9cal12y9tzwfqt7bkxoqkjd2iuz0n3.png)
So, all together: (same denominator; combine terms)
![(1)/(cos^2(x))-(sin^2(x))/(cos^2(x))=(1-sin^2(x))/(cos^2(x))](https://img.qammunity.org/2021/formulas/mathematics/middle-school/qkhx6uehthc0a2k794xd4i2hhbi9n6xg2n.png)
Note the numerator; it can be derived from the Pythagorean Identity:
![sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/pgk4dz6t4trmuf26c7to05t1vnsf4pxt53.png)
Thus, we can substitute the numerator:
![(1-sin^2(x))/(cos^2(x))=(cos^2(x))/(cos^2(x))=1](https://img.qammunity.org/2021/formulas/mathematics/middle-school/qfqb4xbtopuhsfyw8orzjnjfltvcdhaak5.png)
Everything simplifies to 1.