Question

@DanJS

Answer 1

The first one is
\(\bf cot~x~sec^4x~=~cot~x~+~2~tan~x~+~tan^3x\\\)

Answer 2

ok

Answer 3

give me a sec to type...

Answer 4

ok

Answer 5

\[\frac{ \cos x }{ \sin x} *\frac{ 1 }{ \cos^4 x } = \frac{ \cos x }{ \sin x } + 2\frac{ \sin x }{ \cos x } + \frac{ \sin^3 x }{ \cos^3 x }\]

Answer 6

u know the basic definition cot = cos / sin ....and tan = sin/cos ?

Answer 7

yes

Answer 8

Multiplying everything by the common denominator of:

\[\frac{ \sin x \cos^4 x }{ \sin x \cos^4 x }\]

Answer 9

\[\frac{ \cos x }{ \sin x \cos^4 x } = \frac{ \cos x \cos^4 x }{ \sin x \cos^4 x } + \frac{ 2\sin^2 x \cos^3 x }{ \sin x \cos^4 x }+ \frac{ \sin^4 x \cos x }{ \sin x \cos^4 x }\]

Answer 10

multiply everything by sin(x)cos^4(x)

\[\cos x = \cos^5 x + 2\sin^2 x \cos^3 x + \sin^4 x \cos x\]

Answer 11

Divide now by cos(x)

\[1 = \cos^4 x + 2\sin^2x \cos^2x + \sin^4 x\]

Answer 12

U have a perfect square here in the form (a+b)^2

\[1 = [\cos^2(x) +\sin^2(x)]^2\]

Answer 13

recall: \[\sin^2(x) + \cos^2(x) =1\]

Answer 14

so you have

1 = 1

TRUE

Answer 15

Ok, I think I get his one a little better now.

Answer 16

Thanks for explaining.

Answer 17

yep, there may be some identity i missed that would make this shorter, not sure, but this works too.