Question

Help me verify this identity.Follow the structure from the notes by identify the properties used in each step.

Answer 1

Given that you have to verify that:

$\frac{1}{1-sinx}+\frac{1-sinx}{cos^2x}=2sec^2x$

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:

$cos^2x+sin^2x=1$

If you solve for:

$cos^2x$

You get:

$cos^2x=1-sin^2x$

Then, you can rewrite the expression on the right side in this form:

$\frac{1}{1-sinx}+\frac{1-sinx}{1-sin^2x}=2sec^2x$

2. By definition:

$(a-b)(a+b)=a^2-b^2$

Then, you can rewrite the denominator of the second fraction in this form:

$\frac{1}{1-sinx}+\frac{1-sinx}{(1-sinx)(1+sinx)}=2sec^2x$

3. Now you can cancel common terms:

$\frac{1}{1-sinx}+\frac{1}{1+sinx}=2sec^2x$

4. Add the fractions using this formula for adding fractions with different denominators:

$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{ad}$

Then, you get:

$\frac{(1)(1+sinx)+(1)(1-sinx)}{(1-sinx)(1+sinx)}=2sec^2x$$\frac{1+sinx+1-sinx}{(1-sinx)(1+sinx)}=2sec^2x$$\frac{2}{1-sin^2x}=2sec^2x$

5. You already know that:

$cos^2x=1-sin^2x$

Then, you can rewrite the denominator as:

$\frac{2}{cos^2x}=2sec^2x$

6. Remember this Reciprocal Identify:

$\frac{1}{cosx}=secx$

Therefore, you can determine:

$2sec^2x=2sec^2x$

Hence, the answer is: