anonymous 2 years ago Use basic identities to simplify the expression. one divided by cotangent of theta to the second power. + sec θ cos θ

1. anonymous

2. hartnn

so its $$\huge \dfrac{1}{\cot^2 \theta + \sec \theta \cos \theta}$$

3. hartnn

firstly, use the identity that sec theta = 1/cos theta so, sec theta cos theta = 1 and do you know what is $$\large 1+\cot^2 \theta = ... ?$$

4. anonymous

only the cot^2theta is the denominator

5. anonymous

no I don't know, I am COMPLETELY lost in this subject

6. hartnn

okk., so anyways you can use sec theta cos theta = 1 $$\large \dfrac{1}{\cot^2 \theta }+1$$

7. hartnn

now 1/ cot theta = tan theta so, $$\large \tan^2 \theta+1$$

8. hartnn

and ther's a standard pythagorean identity that $$\large 1+\tan^2 \theta = \sec^2 \theta$$ thats it! :)

9. anonymous

okay thank you!

10. anonymous

can you walk me through another please?

11. hartnn

sure

12. anonymous

Simplify the expression. quantity cosecant of x to the power of two times secant of x to the power of two divided by quantity secant of x to the power of two plus cosecant of x to the power of two.

13. anonymous

|dw:1389469428352:dw|

14. anonymous

yes

15. hartnn

$$\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x }$$

16. hartnn

ok, we use the identity that csc x = 1/ sin x sec x = 1/cos x

17. anonymous

okay

18. hartnn

$$\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x } = \large \dfrac{\sin^2 x \cos^2 x }{\sin^2 x + \cos^2 x }$$

19. hartnn

now we can use the famous pythagorean identity $$\Large \color{red}{\sin^2x + \cos^2 x =1 }$$

20. hartnn

21. anonymous

so the answer is just 1? or am I wrong?

22. hartnn

denominator = 1 yes what remains is numerator sin^2 x cos^2 x

23. hartnn

you have choices for this too ?

24. anonymous

yes I do the choices are: sin^2x cos^2x -1 1

25. anonymous

im sorry im so stupid. im usually really good at math but precalc just makes no sense to me

26. hartnn

oops, i did a simplification error $$\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x } = \large \dfrac{1 }{\sin^2 x + \cos^2 x } = 1$$

27. anonymous

okay and i have a question. next year I planned on taking ap stat. is that anything like precalc because i hope not. is it easier or harder?

28. hartnn

if you practice less, math is hard. if you practice more, math is easy. irrespective of any topic/subject

29. anonymous

well are the concepts of ap stat very similar to or very different to those of pre-calculus?

30. hartnn

statistics is a different branch of math altogether, so very different.

31. anonymous

i just think I'm not good at online class taking because I learn better from an actual person rather than a computer screen. I need the physical classroom environment and a teacher to give me instruction.

32. anonymous

what kind of problems does ap stat entail? word problems? graphs?

33. hartnn

both.

34. anonymous

oh okay thank you soooooo much! is there anything else that i should know?

35. hartnn

try to take up easy questions before going to harder ones. if you get stuck anytime, OpenStudy is always Open for you :)