Question

Yes
Yes
No
No
Yes

Answer 1

good

Answer 2

Im having trouble with the reasoning

Answer 3

well let's go one by one

1. cos^2(x)(1+tan^2(x))

if we re-write tan in terms of sin and cos

cos^2(x) * (1 + sin^2/cos^2x)

distributing gives us

cos^2(x) + sin^2(x) = 1

^ done, proved.

Answer 4

2. sec(x)tan(x)(1-sin^2(x))

similar reasoning here, re-write tanx as sinx/cosx and sec(x) as 1/cos(x)

(1/cos(x)) * sin(x)/cos(x) * (1 - sin^2(x))

distribute

sin(x)/cos^2(x) - sin^3(x)/cos^2(x)

combining these into one expression

sin(x) - sin^3(x) / cos^2(x)

--> let me draw this out so it'll be a little easier to see

Answer 5

3 is not an identity, we can attempt to disprove this by re-writing csc as 1/sin and distributing

(1/sinx)(2sin(x) - sqrt(2)

= 2 - sqrt(2)/sin(x)

there's no reason for this to be equal to 0 under all circumstances thus disproving the statement

Answer 6

4 is not an identity either, using the identity

cos(2x) = cos^2(x) - sin^2(x)

we can apply this to the original problem to get

cos^2(2x) - sin^2(4x) = cos(4x)

which is not equal to 0 under all circumstances thus disproving the identity

Answer 7

5 is easy, if we re-write csc^2 as 1/sin^2(x) the left side cancels out to 1 which gives us the real identity cos^2(x) + sin^2(x) = 1

QED

Answer 8

anyhoo that's a lot to take in but lmk if something was unclear

Answer 9

wow tthat is a alot to take in but i thin k i get it