Hi, I really need help with these questions. I did some of them halfway, but then I got stuck. Would you please help me? Thank you so much.

Question

Prove the identity....

1. sec x + tan x(1-sin x/cos x)=1

1/cos x + sin x/cos x(cos^2 x/cos x)=1
1+sin x/cos x(cos^2x/cos x)=1
I got stuck right here.

2. 2/sqrt3 cos x +sin x=sec(pi/6-x)
I did not know how to do this.

Please help me. Thank you so much.
one more question...

Prove tan(ᶿ/2)=sinᶿ/1+cosᶿ) for ᶿ in quadrant 1 by filling in the reasons below.

Answer 1

First of all, to prove identities you have to work on either the left side (LS) or the right side(RS) independent of each other
Secondly, ....

1.
For your second line to be valid, the original should have been
sec x + tan x( (1-sin^2 x)/cos x)=1

LS = 1/cos x + sin x/cos x(cos^2 x/cos x)
= 1/cosx + sinx/cosx (cosx)
= 1/cosx + sinx
≠ 1
≠ RS

In most cases, we should assume that "they" have given us a valid identity, that is, the statement is actually true.
Since I could not prove LS = RS, and I can't seem to find my error, I then try any value of x using my calculator.
(it should work for all values)
e.g. try x = 30°
LS = sec30 + tan30(1-tan3) , the way you typed it
= 2/√3 + (1/√3)(1 - 1/√3)
≠ 1

All I need is a single case where the identify fails,
no wonder we could not prove it.

Answer 2

2. Again, I think you have a typo, and meant

2/(√3cosx + sinx) = sec(π/6 - x)

RS = 1/cos(π/6-x) )
= 1/[ cos(π/6)cosx + sin(π/6)sinx ]
= 1/[ (√3/2)cosx + (1/2)sinx ]
= 1/( (1/2)(√3cosx + sinx) )
= 2/(√3cosx + sinx)
= RS , if you had typed it correctly

Can you see how important those brackets are ??

Answer 3

let Ø/2 = x for easier typing, then I should prove

tanx = sin(2x)/( 1+ cos(2x))

RS = 2sinxcosx/(1 + 2cos^2 x - 1)
= 2sinxcosx/(2cos^2 x)
= sinx/cosx
= tanx
= LS

Answer 4

Thank you very much.