Help with trig identity problems

i am so stuck on these problems. any help would be much appreciated. even just helping me figure out how to get started and get through some of the harder stuff would be great. i just hate these and i have the hardest time. im not just trying to get free answers to my homework, i just have a hard time with these and i just need some extra help understanding them. anyway, here are the ones im stuck on:

1. 1-(cos^2x / 1+sinx) = sinx

2. (secx-cscx)/(secx+cscx) = (tanx-1)/(tanx+1)

3. cot^2 2x+cos^2 2x+sin^2 2x = csc^2 2x

4. (secx-tanx)^2 = (1-sinx)/(1+sinx)

5. (cos^2t+4cost+4)/(cost+20 = (2sect+1)/(sect)

6. ((sinx+cosx)/(sinx))-((cosx-sinx)/(cosx)… = secxcscx


Thank you!

1. 1-(cos^2x / 1+sinx) = sinx

ans:
1-(cos^2x / 1+sinx)
=1-((1-sin^x))/1+sinx)
now 1-sin^x can be written as (1-sinx)(1+sinx)
thus
=1-(((1-sinx)(1+sinx))/1+sinx)
cancel the term 1+sinx from numerator & denominator
=1-(1-sinx)
=sinx
hence proved
:)


2. (secx-cscx)/(secx+cscx) = (tanx-1)/(tanx+1)
ans:
now convert the lhs in terms of cos & sines
i.e secx=1/cosx
cosecx=1/sinx
thus we get
(1/cosx-1/sinx)/(1/cosx+1/sinx)
now tk lcm
thus
((sinx-cosx)/sinxcosx)/((sinx+cosx)/(si…
canceling out sinxcosx
(sinx-cosx)/(sinx+cosx)
now divide each term wth cosx
thus
tanx-1/tanx+1
thus proved
:)


3. cot^2 2x+cos^2 2x+sin^2 2x = csc^2 2x
ans;
here we know that cos^2 2x=1-sin^2 2x
thus rhs
= cot^2 2x+1-sin^2 2x+sin^2 2x
=cot^2 2x+1
=cos^2 2x/sin^2 2x +1
tk lcm
=cos^2 2x+sin^2 2x)/sin^2 2x
=1/sin^2 2x
=cosec^2 2x
hence proved:


4. (secx-tanx)^2 = (1-sinx)/(1+sinx)
ans:
(secx-tanx)^2
=sec^2 x -2*secx*tanx +tan^2 x
=1/cos^2 x -2*1/cosx* sin x/cos x + sin^2 x/cos^2 x
tkn lcm cos^2 x
=(1-2sinx +sin^2 x)/cos^2 x
now (1-2sinx +sin^2 x)=(1-sinx)^2
thus eqn bcms
=(1-sinx)^2 / (1-sin^2 x)...................... since cos^2 x=1-sin^2 x
=(1-sinx)(1-sinx)/(1-sinx)(1+sinx)
canclng (1-sinx)
=(1-sinx)/(1+sinx)
hence proved


5. (cos^2t+4cost+4)/(cost+2) = (2sect+1)/(sect)
ans:
now (cos^2t+4cost+4)=(cost + 2)^2
thus rhs cn be
(cost + 2)^2/(cost + 2)
=cost + 2
=1/sect + 2
on taking lcm we get
=(1 + 2sect)/sect
thus proved


6. ((sinx+cosx)/(sinx))-((cosx-sinx)/(cosx)… = secxcscx
ans:
tak lcm sinxcosx
thus
=((sinx+cosx)cosx/(sinxcosx))-((cosx-si…
=[(sinxcosx + cos^2 x)-(sinxcosx-sin^2 x)]/sinxcosx
in num we get sinxcosx-sinxcosx + cos^2 x + sin^2 x
thus eqn becms
=1/sinxcosx
=secxcosecx
thus proved

hope you cn do ur hmwrk now....
i hope u r clear