# Rep For Help, Maths Identities

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#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
17 years ago
#2
a)

LHS == sec² + cosec²
== 1/cos² + 1/sin²
== sin²/(cos²sin²) + cos²/(sin²cos²)
== (sin² + cos²)/(sin²cos²)
== 1/(sin²cos²)
== sec²cosec²
== RHS
0
17 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
17 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
17 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
17 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
17 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
17 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
#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
17 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
17 years ago
#11
(Original post by elpaw)
alt+0178 = ² and alt+0179 = ³
how do u get the integrate one?
0
17 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
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