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Solving Trig Identities (w/ fractions, poss. work on both sides)?
Okay, I'm lost and frustrated at this point. I seem to be one, inspiration, or step away from solving these type of equations, but it eludes me. The simpler ones, no prob, but when it comes to the following equations, I am foooked:
(-1/[tan - sec) + (-1[tan + sec]) = 2tan
(tan/[1 + cos]) + (sin[1- cos] = cot + sec*csc
(1 - sin)/(1 + sin) = sec^2 - 2sec*tan + tan^
If you know the answer, can you please, pleaseeeee show me all the steps and give me a little outline on your method? I need to see what steps I am not doing correctly or at all.
Verifying, checking, not solving, sorry. :D
1 Answer
- mohanrao dLv 71 decade agoFavorite Answer
-1.................-1
_______+ ________
tan - sec...(tan + sec)
taking common Denominator
[ -tan - sec - tan + sec ]
_____________________
(tan^2 - sec^2)
-2tan/(-1)
= 2tan
2)
tan/(1+cos) + sin/(1 - cos)
= [tan(1 - cos) + sin(1 + cos)]/(1 - cos^2)
= [tan - tan cos + sin + sin cos ]/ sin^2
= [ (sin/cos) - sin + sin + sin cos ] /sin^2
= [sin + sin cos^2 ] /sin^2 cos
= (1/sin cos ) + (cos /sin )
= cot + csc sec
3)
(1 - sin)/(1 + sin)
= (1 - sin)^2 /(1 - sin^2 )
= [1 + sin^2 - 2sin ] /cos^2
= sec^2 + tan^2 - 2 tan sec
