# Rep For Help, Maths Identities

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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) (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

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#2

a)

LHS == sec² + cosec²

== 1/cos² + 1/sin²

== sin²/(cos²sin²) + cos²/(sin²cos²)

== (sin² + cos²)/(sin²cos²)

== 1/(sin²cos²)

== sec²cosec²

== RHS

LHS == sec² + cosec²

== 1/cos² + 1/sin²

== sin²/(cos²sin²) + cos²/(sin²cos²)

== (sin² + cos²)/(sin²cos²)

== 1/(sin²cos²)

== sec²cosec²

== RHS

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#3

(Original post by

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

**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

= (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

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#4

**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

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#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...

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...

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#6

(Original post by

a)

LHS == sec² + cosec²

== 1/cos² + 1/sin²

== sin²/(cos²sin²) + cos²/(sin²cos²)

== (sin² + cos²)/(sin²cos²)

== 1/(sin²cos²)

== sec²cosec²

== RHS

**elpaw**)a)

LHS == sec² + cosec²

== 1/cos² + 1/sin²

== sin²/(cos²sin²) + cos²/(sin²cos²)

== (sin² + cos²)/(sin²cos²)

== 1/(sin²cos²)

== sec²cosec²

== RHS

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#7

(Original post by

Elpaw, how do you get the "squared" and "cubed" characters?

**bono**)Elpaw, how do you get the "squared" and "cubed" characters?

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#8

**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

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:-)

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(Original post by

Isn't it kinda pathetic that people have to bribe with rep for help?

**capslock**)Isn't it kinda pathetic that people have to bribe with rep for help?

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#10

(Original post by

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?

**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?

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#12

(Original post by

how do u get the integrate one?

**lgs98jonee**)how do u get the integrate one?

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