Help!! Grade 12 Math!!! By Tomorrow?

Question

plz show the steps plz i need your help then free 10 pts
prove the identity or i cannot attend college plz im desperate for help if you answer these correctly then i can enter plz help me
1) (1+sin x) / cos x + cos x / (1+ sin x) = 2 sec x

2) (sinx +tan x) / (1+ cos x) = 1/ cot x

3) sec ^2 x / ((sec^2 x) - 1) = csc ^2 x

4) sincos2+sin^3x = 1 / csc x

Answer 1

1) (1+sin x) / cos x + cos x / (1+ sin x) =
(1+sinx)² + cos²x /cosx*(1+sinx)=
(1+2*1*sinx+ sin²x + cos²x) /cosx*(1+sinx)=

As sin²x+cos²x = 1:

(1+2*sinx+ 1) /cosx*(1+sinx)=
(2*sinx+ 2) /cosx*(1+sinx)=
2*(sinx+ 1) /cosx*(1+sinx)=
2 /cosx=
2*(1/cos x) =
2 sec x

2) (sinx +tan x) / (1+ cos x) = 1/ cot x
(sinx +tan x) / (1+ cos x) =
(sinx +(sinx/cosx)) / (1+ cos x) =
((sinx*cosx +sinx)/cosx) / (1+ cos x) =
((sinx*cosx +sinx)/cosx) *1/(1+ cos x) =
(sinx*cosx +sinx)/ (cosx*(1+cosx)) =
(sinx*(cosx +1))/ (cosx*(1+cosx)) =
sinx/ cosx =

As cotx = cos x / senx:

1/cot x


3) sec ^2 x / ((sec^2 x) - 1) = csc ^2 x
sec ^2 x / ((sec^2 x) - 1) =

sin²x +cos²x = 1

Dividing by cos²x:

sin²x/cos²x +cos²x/cos²x = 1/cos²x
tan²x + 1 = sec²x
tan²x = sec²x -1


sec ^2 x / ((sec^2 x) - 1) =
sec²x / tg²x =
(1/cos²x) / (sin²x/cos²x) =
(1/cos²x) * (cos²x/sin²x) =
(1/sin²x) =
csc²x

4) sinx[cos^2x]+sin^3x =
sinx (cos²x+sin²x) =
sinx*1 =
sinx

As csc x = 1/sin x:

=1/csc x

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=**

Answer 2

1) multiply the whole equation by cos x (1+sin x) to get:

(1+sin x)^2 + (cos x)^2 = 2 (cos x (1+sin x)/ cos x)
1 + 2sin x + (sin x)^2 + (cos x)^2 = 2(1+sinx)
1 + 2sin x + (sin x)^2 + 1-(sin x)^2 = 2(1+sinx)
2 + 2sin x = 2(1+sinx)
2(1 + sin x ) = 2(1+sinx)
QED

3) sec ^2 x / ((sec^2 x) - 1) = csc ^2 x

(1/cos²x ) / ((1/cos²x )-1) = 1/sin²x
(1/cos²x ) / ((1-cos²x )/cos²x) = 1/sin²x
using sin²x+cos²x = 1 OR sin²x= 1-cos²x
(1/cos²x ) / ((sin²x )/cos²x) = 1/sin²x = 1/sin²x
QED

2) (sinx +tan x) / (1+ cos x) = 1/ cot x

[sin x + sinx / cosx] / (1+ cos x) = sin x / cosx

[sinxcosx + sinx]/[cosx (1+ cos x)]= sin x / cosx
sinx(1+cosx)/[cosx (1+ cos x)]= sin x / cosx = sin x / cosx
QED

4) sinxcos²x+sin³ x= 1 / csc x = sinx
factor out sinx and use sin²x+cos²x = 1 to get
sinx(cos²x+sin²x)= sinx (1) =sinx = left side sinx
QED

Answer 3

For future - look here: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

s=sin x
c=cos x
t=tan x
cot=cot x
sec=sec x
csc = csc x

1)
(1+s)/c + c/(1+s)
= [(1+s)^2 +c^2]/c(1+s)
= [1+2s +s^2 +c^2]/c(1+s) use s^2+c^2=1
= 2(1+s)/c(1+s)
=2/c
=2.sec from cos(x)=1/sec(x)

2)
(s+t)/(1+c) use t=s/c
=(s+s/c)/(1+c)
=s(1+1/c)/(1+c)
=s[(c+1)/c]/(1+c)
=s/c
=t use tan(x)=1/cot(x)
=1/cot

3)
sec^2/{sec^x-1) use 1+[tan(x)]^2= [sec(x)]^2
=(1+t^2)/t^2
=1/t^2 +1 use cot=1/t
=cot^2 +1 use csc^2=1+cot^2
=csc^2

4) assuming you mean (sin x)(cos x)^2 +(sin x)^3
s.c^2 +s^3
=s.(c^2+s^2) use c^2 + s^2=1
=s use csc=1/sin
=1/csc

Answer 4

I'll do two of them. That should give you a good idea of how to go about the other two.

2) (sinx + tan x) / (1+ cos x)

= (sin(x) + sin(x)/cos(x)) / (1 + cos(x))
= (sin(x)cos(x)/cos(x) + sin(x)/cos(x)) / (1 + cos(x))
= [(sin(x)cos(x) + sin(x)) / cos(x)] / (1 + cos(x))
= (sin(x)cos(x) + sin(x)) / (cos(x) * (1 + cos(x))
= (sin(x)cos(x) + sin(x)) / (cos(x) + cos^2(x))
= sin(x)[cos(x) + 1] / cos(x)[1 + cos(x)]
= sin(x) / cos(x)
= tan(x)
= 1 / cot(x)

3) sec ^2 x / ((sec^2 x) - 1) = csc ^2 x

sec ^2 x / ((sec^2 x) - 1)
= sec^2(x) / tan^2(x)
= [1 / cos^2(x)] / [sin^2(x) / cos^2(x)]
= cos^2(x) / cos^2(x)sin^2(x)
= 1 / sin^2(x)
= csc^2(x)

Answer 5

I'll prove the one's I can first.

4. sincos^2(x) + sin^3(x) = 1/csc x

1/csc x = sin x

sin cos^2 x + sin^3 x = sin x

Divide by sin x

cos^2 x + sin^2 x = 1

Don't think I need to explain any further. It's something you should remember.

3. sec^2 x / (sec^2 x - 1) = csc^2 x

sec^2 x -1 = tan^2 x

sec^2 x / tan^2 x = csc^2 x

We can cancel out all the squares.

sec x / tan x = csc x

(hyp / adj)(adj / opp) = (hyp / opp)

If you can out everything on the left side:

hyp / opp = hyp / opp

Thus the identity is true.

2. (sin x + tan x) / (1 + cos x) = 1/cot x

cot x = tan x

(sin x + tan x) / (1 + cos x) = tan x

Multiply (1 + cos x) on both sides.

sin x + tan x = tan x + (tan x) (cos x)

Subtract tan on both sides.

sin x = (tan x)(cos x)

opp / hyp = (opp/adj)(add/hyp)

opp / hyp = opp / hyp

sin x = sin x

So this is true too.

1. (1 + sin x) / cos x + cos x / (1 + sin x) = 2sec x

Multiply cos x on both sides.

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

Multiply (1 + sin x) on both sides.

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

Cancel out things.

sin^2 x + cos^2 x = 1

Answer 6

First of all calm down and take deep breathes. Go over your math book and do some practive questions, to see if you understand the lesson. Thats all you can do. Try not to think of it til the test in math. Remember dont worry!

Answer 7

try doing your own homework. maybe you shouldnt be in these classes

Answer 8

1.1+sinx/cosx+cosx/(1+sinx)
1+sinx/cosx+cosx(1-sinx)/(1-sin^2x)
=(1+sinx)/cosx+(1-sinx)/cosx
=2/cosx
= 2sec x.QED
2.(sinx+tanx)/(1+cosx)
=(sinx+sinx/cosx)/(1+cosx)
=sinx/cosx(cosx+1)/(1+cosx)
=tanx
=1/cotx.QED
3.sec^2x/(sec^2x-1)
=sec^2x/tan^2x
=sec^2x/ sin^2x/cos^2x
= sec^2x cos^2x/sin^2x
=1/sin^2x
= cosec^2x QED.
4.sinx[cos^2x]+sin^3x
= sinx[cos^2x+sin^2x]
=sinx x1
=1/cosec x QED.
=
=

Answer 9

how is it that your college career depends on trig problems?