Can you draw the

Question

The solution for this:

Establishing identities
1.) Sin²∅ (1+cot²∅) = 1

2.) (tan²B+1) cos²B = 1

3.) tan x
---------- = sin x
sec x

4.) 1 1
------- + ------- = 2csc²∅
1+cos∅ 1-cos

Please help me, i cant understand it.

Answer 1

Most of these appear to be based on the Pythagorean identity
sin^2 x + cos^2 x = 1 , and its variations

e.g. if I divide the above by sin^2 x , I get
1 + cot^2 x = csc^2 x
and that is what I see in the LS of #1

Use LS for left side and RS for right side, work each side independently . I usually start with the more complicated looking side and try to change it to the expression on the other side.
Sometimes you have to work on both sides.
As long as you reach some point where both sides are the same, you have done your job.

so #1
LS = sin^2 Ø (1+cot^2 Ø)
= sin^2 Ø (csc^2 Ø)
= 1
= RS

#2, just one of the other variations, you try it

#3. just based on basic definitions,
LS = (sinx/cosx) / (1/cosx)
= (sinx/cosx)(cosx/1)
= sinx
= RS

#4.

LS = 1/(1+cosx) + 1/(1-cosx)
form a common denominator and simplify
= (1 - cosx + 1 + cosx)/(sin^2 x)
= 2/sin^2 x
= 2csc^2 x
= RS

Answer 2

One more clue.
Since all 6 trig functions can be defined in terms of sin and cos, I often change everything to sines and cosines, unless I see an obvious definition such as
1 + cot^2 Ø = csc^2 Ø
in #1

Answer 3

How it looks like , ? #1 to get this 1 + cot^2 x = csc^2 x ??

Answer 4

I get it!