This discussion is closed.
GH
Badges: 13
Rep:
?
#1
Report Thread starter 16 years ago
#1
Prove the identity:

a) (sec^2 theta) + (cosec^2 theta) = (sec^2 theta)*(cosec^2 theta)

b) (1-sin theta)/(1+sin theta) = (sec theta - tan theta)^2

Rep for help as always
0
elpaw
Badges: 15
Rep:
?
#2
Report 16 years ago
#2
a)

LHS == sec² + cosec²
== 1/cos² + 1/sin²
== sin²/(cos²sin²) + cos²/(sin²cos²)
== (sin² + cos²)/(sin²cos²)
== 1/(sin²cos²)
== sec²cosec²
== RHS
0
It'sPhil...
Badges: 2
Rep:
?
#3
Report 16 years ago
#3
(Original post by 2776)
Prove the identity:

a) (sec^2 theta) + (cosec^2 theta) = (sec^2 theta)*(cosec^2 theta)

b) (1-sin theta)/(1+sin theta) = (sec theta - tan theta)^2

Rep for help as always
a) LHS = 1/cos^2x + 1/sin^2x
= (sin^2x + cos^2x)/(sin^2x.cos^2x)
= 1/(sin^2x.cos^2x)
= sec^2x.cosec^2x = LHS

b) LHS = (1-sinx)^2/(1+sinx)(1-sinx)
= ((1-sinx)/cosx)^2
= (secx - tanx)^2 = RHS
0
capslock
Badges: 1
Rep:
?
#4
Report 16 years ago
#4
(Original post by 2776)
Prove the identity:

a) (sec^2 theta) + (cosec^2 theta) = (sec^2 theta)*(cosec^2 theta)

b) (1-sin theta)/(1+sin theta) = (sec theta - tan theta)^2

Rep for help as always
Isn't it kinda pathetic that people have to bribe with rep for help?
0
Saichu
Badges: 9
#5
Report 16 years ago
#5
a. sec^2 + csc^2 = sec^2*csc^2
1/cos^2 + 1/sin^2 = 1/(cos^2*sin^2)
(cos^2 + sin^2)/(sin^2*cos^2) = 1/(cos^2*sin^2)
1/(cos^2*sin^2) = 1/(cos^2*sin^2)

b. (1 - sin)/(1 + sin) = (sec - tan)^2
(1 - sin)/(1 + sin) = sec^2 - 2sectan + tan^2
1 - sin = sec^2 - 2sectan + tan^2 + sin (sec^2 - 2sectan + tan^2)
1 - sin = sec^2 - 2sectan + tan^2 + tan sec - 2tan^2 + sintan^2
1 - sin = sec^2 - sectan - tan^2 + sintan^2
1 - sin = sec^2 - sec^2*sin - sec^2*sin^2 + sec^2*sin^3
1 - sin = sec^2(1 - sin^2 - sin+ sin^3)
1 - sin = sec^2(cos^2 - sin(1 - sin^2))
1 - sin = sec^2(cos^2 - sin*cos^2)
1 - sin = 1 - sin

Edit: Yeh, it would probably have been easier to square it...
0
username9816
Badges: 12
Rep:
?
#6
Report 16 years ago
#6
(Original post by elpaw)
a)

LHS == sec² + cosec²
== 1/cos² + 1/sin²
== sin²/(cos²sin²) + cos²/(sin²cos²)
== (sin² + cos²)/(sin²cos²)
== 1/(sin²cos²)
== sec²cosec²
== RHS
Elpaw, how do you get the "squared" and "cubed" characters?
0
elpaw
Badges: 15
Rep:
?
#7
Report 16 years ago
#7
(Original post by bono)
Elpaw, how do you get the "squared" and "cubed" characters?
alt+0178 = ² and alt+0179 = ³
0
lgs98jonee
Badges: 2
Rep:
?
#8
Report 16 years ago
#8
(Original post by 2776)
Prove the identity:

a) (sec^2 theta) + (cosec^2 theta) = (sec^2 theta)*(cosec^2 theta)

b) (1-sin theta)/(1+sin theta) = (sec theta - tan theta)^2

Rep for help as always
a.1/cos^2+1/sin^2=(sin^2+cos^2)/(sin^2cos^2)
and as sin^2+cos^2=1, it equals 1/(sin^2cos^2)

b. (RHS) sec^2-2tansec+tan^2=1/cos^2-2sin/cos^2+sin^2/cos^2
=(1-2sin+sin^2)/cos^2=(1-sin)^2/cos^2
and as cos^2=1-sin^2
=(1-sin)^2/1-sin^2
=(1-sin)/(1+sin)

yay just checking i can do these without book as have just done them on p2:-)
0
GH
Badges: 13
Rep:
?
#9
Report Thread starter 16 years ago
#9
(Original post by capslock)
Isn't it kinda pathetic that people have to bribe with rep for help?
It would be, but this is for my EDUCATION. Rep to me is worthless without being used, and why not use it so that you can get useful help?
0
username9816
Badges: 12
Rep:
?
#10
Report 16 years ago
#10
(Original post by 2776)
It would be, but this is for my EDUCATION. Rep to me is worthless without being used, and why not use it so that you can get useful help?
People like Elpaw and theone and many others will help you anyway, so would I if I was older
0
lgs98jonee
Badges: 2
Rep:
?
#11
Report 16 years ago
#11
(Original post by elpaw)
alt+0178 = ² and alt+0179 = ³
how do u get the integrate one?
0
elpaw
Badges: 15
Rep:
?
#12
Report 16 years ago
#12
(Original post by lgs98jonee)
how do u get the integrate one?
you just have to find it in the character map, there is no alt+code for it. its unicode is 222B
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

How will you be receiving your results?

Going into school to pick them up (196)
33.39%
Receiving them online / by email (291)
49.57%
I still don't know (100)
17.04%

Watched Threads

View All