trig identities? easy 10 points!!!?

Question

this should be SOO simple and i therefore am guessing that there is a glitch in the answers cause I dont see where i went wrong...maybe, i dunno...
Given that
sin theta = y /r
cos theta = x/r
tan theta = y/x
PROVE the following identity:
[ (1 + cot^2 theta) / (csc^2theta) ] = 1

can anyone help?? i cant get the LS to equal the RS :S
also, a slightly more difficult identity, i was having trouble:

prove:
[ (csc theta) / (1 + csc theta)] + [ (csc theta) / (1 - csc theta)]
= [(2 sin theta) / cos^2 theta)]

please help...i dont know where i am going wrong!! i would appreciate anyone's help! thanks ^_^

Answer 1

Question 1
cos θ = x / r
sin θ = y / r
tan θ = y / x

N = 1 + cot ² θ
N = 1 + cos ² θ / sin ² θ
N = (sin ² θ + cos ² θ) / sin ² θ
N = 1 / sin ² θ


D = cosec ² θ
D = 1/ sin ² θ

N/D = (1/sin ² θ) x (sin ² θ/1) = 1

Question 2
cosec θ(1 - cosecθ) + cosecθ(1 + cosecθ)
becomes N / D where:-
N = cosec θ(1 - cosec θ) + cosec θ(1 + cosec θ)
N = 2 cosec θ
N = 2 / sin θ

D = (1 + cosec θ)(1 - cosec θ)
D = 1 - cosec ² θ
D = 1 - 1 / sin²θ
D= (sin ² θ - 1) / sin ² θ

N/D = [ 2 / sin θ ] [ sin ² θ / (sin ² θ - 1) ]
N/D = 2 sin θ / (- cos ² θ)
N/D = - 2 tan θ sec θ

Answer 2

use the fact that csc =1/sin = r/y and cot = 1/tan=x/y

[ (1 + cot^2 theta) / (csc^2theta) ] = 1
[ (1+(x/y)^2 ) / (r/y)^2 ] = 1
Separate into two fractions
[ 1 / (r/y)^2 ] + [ (x/y)^2 / (r/y)^2 ] =1
[ 1 / (r/y)^2 ] + [ (x^2/y^2) * (y^2/r^2) ] =1
[ (y/r)^2 ] + [ (x/r)^2] = 1
Look familiar? Replace y/r and x/r with sin and cos
sin^2 theta + cos^2 theta = 1 which is an identity.
1=1

2. I didn't actually get to an answer but I'll show what I started and hopefully it'll help someone else.

[ (csc theta) / (1 + csc theta)] + [ (csc theta) / (1 - csc theta)]
= [(2 sin theta) / cos^2 theta)]

[ (r/y) / (1+(r/y)) ] + [ (r/y)/(1-(r/y)) ] = [ (2y/r) / (x/r)^2 ]
Let's simplify the LS first
Add the fractions on the left by finding the LCD - which is (1-(r/y)^2 )
[ (r/y)*(1-(r/y)) + (r/y)*(1+(r/y)) ] / (1-(r/y)^2)
[ r/y - (r/y)^2 + (r/y) + (r/y)^2 ] / (1-(r/y)^2 )
2(r/y) / (1-(r/y)^2)

Answer 3

I don't know if you can use my answer since I don't use the given formulas...

but...

using pythagorean identities:
1 + cot^2theta = csc^2theta

therefore:
csc^2 theta / csc^2 theta = 1

I mean, you could go backwards from there, plugging in the given variables into your solution, or just prove that the pythagorean identity is true.

The following is a list of trig identities if you need any more help

Answer 4

well your first problem is really easy

you listed:
sin theta = y /r
cos theta = x/r
tan theta = y/x
all of these doenst matter and you dont need it
if you look at the trig identities it shows that (1 =cot^2 theta) = csc^2theta so it would leave you with
[ (csc^2theta) / (csc^2theta) ] = 1, which cancel out to equal 1
currenty working on the other problem

Answer 5

First, we bring the csc to the top:

sin^2theta (1 + cot^2theta) = 1

Now multiply through:

sin^2theta + sin^2theta * cot^2theta = 1

remembering cot^2theta = cos^2theta / sin^2theta:

sin^2theta * cot^2theta reduces to cos^2theta

So we're simply left with sin^2theta + cos^2theta = 1.

sin^2theta = y^2/r^2
cos^2theta = x^2/r^2

adding them: you get (x^2 + y^2)/r^2 = 1
x^2 + y^2 = r^2 (as defined by the Pythagorean theorem)
so r^2 / r^2 = 1

--------------
Next one:

[ (csc theta) / (1 + csc theta)] + [ (csc theta) / (1 - csc theta)]
= [(2 sin theta) / cos^2 theta)]

Find the common denom on the left side: so multiply the top and bottom of [ (csc theta) / (1 + csc theta)] by (1-csctheta)

and multiply top and bottom of [ (csc theta) / (1 - csc theta)] by (1+csctheta)

so that turns the left side into:
[csctheta - csc^2theta + csctheta + csc^2theta] / [(1-csctheta)(1+csctheta)]

= [(2 sin theta) / cos^2 theta)]

Simplifying and multiplying the bottom through:
(2csctheta)/(1 - csc^2theta) = [(2 sin theta) / cos^2 theta)]

Now, lets multiply top and bottom of the left side by sin^2theta:
That leaves us with:
(2sintheta)/(sin^2theta - 1) = [(2 sin theta) / cos^2 theta)]

Not sure if i lost a sign somewhere or if the identity is wrong...but the bottom is supposed to simplify to cos^2theta (1 - sin^2theta = cos^2theta)....but hopefully you get the idea :)

Answer 6

I have always been completely horrible with trig functions. They're the worst. What I would do was just write a whole bunch of nonsense for a couple lines then put the two sides equal at the end. It would get me a few points on the test. Good luck!!

Answer 7

(1+cot^2theta)/csc^2theta
=
1+(x/y)^2
-------------
(r/y)^2

=
y^2+x^2
-----------
r^2

but we know from pythagoras that r^2=x^2+y^2

so
y^2+x^2
-----------
r^2

=1

probably need that trick for second part as well

Answer 8

Stop making my head hurt :-)

Answer 9

both sides are OK.