Trig. simplification help

tan(127)+tan(-143)+tan(-143)-tan(-53)
--------------------------------------… * sin^2(757) =
tan(217)sin(683)[cos(217)+cos(143)]
Result: tan(37)


Top: tan(127)+tan(-143)+tan(-143)-tan(-53)
=tan(180-53) +[-tan(90+53)] +[-tan(90+53)] -[-tan(53)]

=tan(53)-tan(53)-tan(53)+tan(53) <-- I end up with -tan(53) and +tan(53). Can I just cancel out the top part? Probably not, that would mean the top becomes 0


Bottom: tan(217)sin(683)[cos(217)+cos(143)]
=tan(180+37)sin(360 +180 +90 +53)[cos(180 +37) +cos(90 +53)]

=tan(37)sin(53)[cos(37) +cos(53)]

Putting together what we got:

tan(53)-tan(53)-tan(53)+tan(53)
--------------------------------------… *sin^2(37)
tan(37)sin(53)[cos(37) +cos(53)]

What next? I know things cancel out and I can simplify further. I probably have to use formulas like tan a=sin a/cos a
cot a =cos a/sin a etc.
PS: What are these formulas called? Where can I find all of them? Cause I can't find them anywhere online, and no I don't have any textbooks.

Thanks in advance!

Top: tan(127) + tan(-143) + tan(-143) - tan(-53)

since tan is odd function, tan(-A) = -tan A

=> tan(127) - tan(143) - tan(143) + tan(53)

=> tan(127) - 2tan(143) + tan(53)

= tan(180 - 53) - 2tan(90 + 53) + tan(53)

recall that tan(90 + A) = - cot A

= - tan(53) + 2cot(53) + tan(53)

= 2 cot(53) ------------TOP

Bottom: tan(217)sin(683)[cos(217)+cos(143)]

= tan(180 + 37)sin(720 - 37)[cos(180 + 37) + cos(180 - 37 ]

= tan(37)(- sin(37)[ - cos(37) - cos(37) ]

= tan(37)sin(37) [2 cos(37) ]

= 2sin^2(37) -----------Bottom

and sin^2( 757) = sin^2(720 + 37)

= sin^2(37)

so final step is

2 cot(53)* sin^2(37)
_________________
2sin^2(37)

= cot(53)

= cot(90 - 37)

recall that cot (90 - A) = tan A

= tan(37)

The formulas you're looking for are called trigonometric identities. Here's a good list of some of the key identities: http://www.sosmath.com/trig/Trig5/trig5/…