proofing of trigo help

[Alexis99]

Below is the half way proof that I have done. I did not know how to continue from here. Can anyone help? Many thanks.

i) Proof that (1-sin2A)/(1+cos2A) = 1/2 (1+tanA)^2

LHS: (1-2sinAcosA)/(1+2cos^2A-1)

= (1-2sinAcosA)/2cos^2A

= 1/2cos^2A - 2sinAcosA/2cos^2A

= 1/2sec^2A - sinA/cosA

= 1/2(tan^2A+1) - tanA


ii) Proof that cosec x (1 + cosx) + 1/cosec(1+cosx) = 2cosecx

LHS: 1/sinx + cos x / sin x + 1/cosec(1+cosx)

= (1 + cosx) /sin x + 1/sin (1+cosx)

 

[Ishuda]

Below is the half way proof that I have done. I did not know how to continue from here. Can anyone help? Many thanks.

i) Proof that (1-sin2A)/(1+cos2A) = 1/2 (1+tanA)^2

LHS: (1-2sinAcosA)/(1+2cos^2A-1)

= (1-2sinAcosA)/2cos^2A

= 1/2cos^2A - 2sinAcosA/2cos^2A

= 1/2sec^2A - sinA/cosA

= 1/2(tan^2A+1) - tanA


ii) Proof that cosec x (1 + cosx) + 1/cosec(1+cosx) = 2cosecx

LHS: 1/sinx + cos x / sin x + 1/cosec(1+cosx)

= (1 + cosx) /sin x + 1/sin (1+cosx)

For (1), to continue you have

= 1/2(tan^2A+1) - tanA
= (1/2) (tan²(A) - 2 tan(A) + 1 )
= (1/2) (tan(A) - 1)²
and the original equation is incorrect. As an example use A = $\displaystyle \frac{\pi}{4}$:
sin(2A) = 1
cos(2A) = 0
tan(A) = 1
and
(1-sin2A)/(1+cos2A) $\displaystyle \ne$ 1/2 (1+tanA)^2

For (2) grouping symbols would help, i.e.
cosecx (1 + cosx) + 1/( cosecx (1+cosx) ) = 2 cosecx
LHS: Since 1/cosecx is sinx, we can write this as
1 / (sinx (1 + cosx) ) + sinx / (1+cosx)
now put everything over a common denominator and simplify.