trig identity!!!!!!!!!!!!!!

Question

Ok, two problems. Need help. 

Prove that

1. tan^(4) theta   -   sec^(4) = 1 - 2 sec^(2) theta 

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

Answer 1

1) LHS   = tan^4θ - sec^4θ

              = tan^4θ - (sec^2θ)^2                                   

              = tan^4θ - (1 + tan^2θ)^2                                  [ Since sec^2θ = 1 + tan^2θ ]

              = tan^4θ - (1+ tan^4θ + 2(1)(tan^2θ))               [ Since (a + b)^2 = a^2 +b^2 + 2ab ]

              = tan^4θ - (1 + tan^4θ + 2tan^2θ)

              = tan^4θ - 1 - tan^4θ - 2tan^2θ

              = -1 - 2tan^2θ

              = -1 - 2(sec^2θ - 1)                                          [ Since sec^2θ = 1 + tan^2θ ]

              = -1 - 2sec^2θ + 2

              = 1 - 2sec^2θ

              = RHS

Hence it is proved that tan^4θ - sec^4θ = 1 - 2sec^2θ

Answer 2

2) LHS = (sinθ + 1) / (sinθ - 1)

          = (sinθ + 1)^2 / (sinθ - 1)(sinθ + 1)             [ Dividing numerator and denominator by (sinθ + 1) ]

          = (sin^2θ + 1^2 + 2sinθ(1)) / (sin^2θ - 1^2) [ Since (a+b)^2=a^2+b^2+2ab, (a-b)(a+b)=a^2-b^2 ]

          = (sin^2θ  + 1 + 2sinθ) / (sin^2θ  - 1)          

          = (sin^2θ  + 1 + 2sinθ) / -cos^2θ                  [ Since sin^2x +cos^2x = 1 ]

          = -sin^2θ/cos^2θ - 1/cos^2θ - 2sinθ/cos^2θ  [ Splitting the terms ]

          = -tan^2θ  - sec^2θ  - 2tanθ secθ                 [ Since sinx/cosx=tanx, 1/cosx=secx ]

          = -(sec^2θ - 1) - sec^2θ -2tanθ secθ            [ Since tan^2x = sec^2x - 1]

          = -sec^2θ +1 -sec^2θ - 2tanθ secθ

          = 1 - 2sec^2θ - 2secθ tanθ

          = RHS

Hence it is proved that (sinθ + 1) / (sinθ - 1) = 1 - 2sec^2θ - 2secθ tanθ