Given that you have to verify that:
![(1)/(1-sinx)+(1-sinx)/(cos^2x)=2sec^2x](https://img.qammunity.org/2023/formulas/mathematics/college/79zwbqea5v8uybxu8u079kuvnu2jrdyege.png)
You need to follow these steps, in order to solve the exercise:
1. You need to use this Pythagorean Identify to rewrite the denominator of the second fraction:
![cos^2x+sin^2x=1](https://img.qammunity.org/2023/formulas/mathematics/college/vz35usfrwxshaaxvqfk2mvgff8uzvo6st6.png)
If you solve for:
![cos^2x](https://img.qammunity.org/2023/formulas/mathematics/college/e8r3x76cpwddlxxt7rxtklak8yygrediu3.png)
You get:
![cos^2x=1-sin^2x](https://img.qammunity.org/2023/formulas/mathematics/college/nfridpkhupejk0ogvnp1ysjf6wrvwdneq1.png)
Then, you can rewrite the expression on the right side in this form:
![(1)/(1-sinx)+(1-sinx)/(1-sin^2x)=2sec^2x](https://img.qammunity.org/2023/formulas/mathematics/college/lac0j9gel38xh070xm6owpiikqm6tcpuic.png)
2. By definition:
![(a-b)(a+b)=a^2-b^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/hjpeozjrpf6sx411hrfrujuyqigmou3r0c.png)
Then, you can rewrite the denominator of the second fraction in this form:
![(1)/(1-sinx)+(1-sinx)/((1-sinx)(1+sinx))=2sec^2x](https://img.qammunity.org/2023/formulas/mathematics/college/a6d3mzpusdfjfpg1fk6658lsif7x94f8bc.png)
3. Now you can cancel common terms:
![(1)/(1-sinx)+(1)/(1+sinx)=2sec^2x](https://img.qammunity.org/2023/formulas/mathematics/college/j4kvt2jmoe8fmakow6fle8gg3pispuxrlm.png)
4. Add the fractions using this formula for adding fractions with different denominators:
![(a)/(b)+(c)/(d)=(ad+bc)/(ad)](https://img.qammunity.org/2023/formulas/mathematics/college/7nd82stb4ubxmno0rvjlmyjurtkibempgi.png)
Then, you get:
![((1)(1+sinx)+(1)(1-sinx))/((1-sinx)(1+sinx))=2sec^2x](https://img.qammunity.org/2023/formulas/mathematics/college/256mglgbvyjswxabb4go0tddocrdtp8cds.png)
![(1+sinx+1-sinx)/((1-sinx)(1+sinx))=2sec^2x](https://img.qammunity.org/2023/formulas/mathematics/college/fcqe7c11jf4z7zxy6wf26v71trifdvo96r.png)
![(2)/(1-sin^2x)=2sec^2x](https://img.qammunity.org/2023/formulas/mathematics/college/euirmg1usphqdkletqd1gy0n23ri6oc7rw.png)
5. You already know that:
![cos^2x=1-sin^2x](https://img.qammunity.org/2023/formulas/mathematics/college/nfridpkhupejk0ogvnp1ysjf6wrvwdneq1.png)
Then, you can rewrite the denominator as:
![(2)/(cos^2x)=2sec^2x](https://img.qammunity.org/2023/formulas/mathematics/college/2syng1jkxepva0ffh09cht7wb0hns6vlgi.png)
6. Remember this Reciprocal Identify:
![(1)/(cosx)=secx](https://img.qammunity.org/2023/formulas/mathematics/college/8ry28576o2u33fql0hh0meiixwzrm6l8k8.png)
Therefore, you can determine:
![2sec^2x=2sec^2x](https://img.qammunity.org/2023/formulas/mathematics/college/1m9olnypiepkl3nmka3it3sfys975deeow.png)
Hence, the answer is: