1. ## Simplifying trigonometric functions

Where do I go on a problem like this? I am supposed to simplify them...I have done about 10 of these earlier today, but I am stuck on these last two...I know that there is probably more than one way to go on these, but I am having a hard time getting started

2. for 66, I have done:
Common denominator

(Sec x - 1) - (sec x + 1)/ (Sec x - 1)(sec x + 1)

3. Originally Posted by aikenfan
for 66, I have done:
Common denominator

(Sec x - 1) + (sec x + 1)/ (Sec x - 1)(sec x + 1)
that is wrong. you were on the right track, but you messed up the signs

$\displaystyle \frac 1{\sec x + 1} - \frac 1{\sec x - 1} = \frac {\sec x - 1 - \sec x - 1}{(\sec x + 1)(\sec x - 1)}$

note that the denominator is the difference of two squares, thus we have:

$\displaystyle \frac {-2}{\sec^2 - 1}$

now what?

continue the same way for the next problem. you don't have many options besides combining the fractions, follow your nose

4. this would make it -2/tan^2x
because of Pythagorean Identities: tan^2x = sec^2x - 1

5. Originally Posted by aikenfan
this would make it -2/tan^2x
because of Pythagorean Identities: tan^2x = sec^2x - 1
yes. but what is 1/tan^2(x) ?

6. cot^2x

7. Originally Posted by aikenfan
cot^2x
yes, so we have -2cot^2(x)

now, move on to the second

8. So far, I have got:
(cosx(1 + sinx) + 1 + sinx(1 + sinx))/(1 + sinx)(cosx)
I am guessing that I have to distribute these

9. Originally Posted by aikenfan
So far, I have got:
(cosx(1 + sinx) + 1 + sinx(1 + sinx))/(1 + sinx)(cosx)
I am guessing that I have to distribute these
yup, that would be a good idea. again, you don't have many other options, so just follow your nose

10. I get:
(cosx + cosx(sinx)) + (1 + sinx + sinx + sinx^2)/ (1 + sinx)(cosx)

= (cosx + cosx(sinx)) + (1 + 2sinx + sinx^2)/ (1 + sinx)(cosx)

11. Originally Posted by aikenfan
I get:
(cosx + cosx(sinx)) + (1 + sinx + sinx + sinx^2)/ (1 + sinx)(cosx)

= (cosx + cosx(sinx)) + (1 + 2sinx + sinx^2)/ (1 + sinx)(cosx)
where did cosx(sinx) come from, you shouldn't have that

12. Hello, aikenfan!

There is a trick we can use on the second problem . . .

$\displaystyle 68)\;\;\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}$
Multiply the first fraction by $\displaystyle \frac{1-\sin x}{1-\sin x}$

. . We have: .$\displaystyle \frac{\cos x}{1 + \sin x}\cdot{\color{blue}\frac{1-\sin x}{1-\sin x}} \;\;=\;\;\frac{\cos x(1 -\sin x)}{1-\sin^2\!x} \;\;=\;\;\frac{\cos x(1-\sin x)}{\cos^2\!x} \;\;=\;\;\frac{1-\sin x}{\cos x}$

The problem becomes: .$\displaystyle \frac{1-\sin x}{\cos x} + \frac{1+\sin x}{\cos x} \;\;=\;\;\frac{2}{\cos x} \;\;=\;\;2\sec x$

13. Thank you both very much for all of your time and help!