Yahoo Answers: Answers and Comments for Prove that tanx + cotx = (secx)(cscx).? [Mathematics]
Copyright © Yahoo! Inc. All rights reserved.
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
From grrrrrrrrrrrrrrrrrrrrrrrrrrrreat
enUS
Sat, 09 Jan 2010 11:49:34 +0000
3
Yahoo Answers: Answers and Comments for Prove that tanx + cotx = (secx)(cscx).? [Mathematics]
292
38
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
https://s.yimg.com/zz/combo?images/emaillogous.png

From maegical: tan x + cot x = sin x/cos x + cos x/sin x =
(...
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
Sat, 09 Jan 2010 11:55:43 +0000
tan x + cot x = sin x/cos x + cos x/sin x =
(sin^2 x/ (cos x *sin x)) + (cos^2 x/(sin x* cos x) =
(sin^2 x + cos^2 x) / (cos x * sin x) =
1/ (cos x * sin x) =
1/ cos x * 1/sin x =
sec x * csc x

From kawaiaea: replace tanx to sinx/cosx then replace cotx to...
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
Thu, 08 Dec 2016 20:38:05 +0000
replace tanx to sinx/cosx then replace cotx to cosx/sinx replace secx to a million/cosx and cscx to a million/sinx sinx/cosx + cosx/sinx=a million/cosx(a million/sinx) while u upload u upload or subtract u could desire to get consumerfriendly denominators for that edge of the equation so situations sinx/cosx by potential of sinx and situations cosx/sinx by potential of cosx sin^2x +cos^2x/cosxsinx=a million/cosx(a million/sinx) sin^2x + cos^2x = a million so replace it for a million a million/cosxsinx=a million/cosx(a million/sinx) so now for the different edge of the equation diverse for the duration of with a million/cosx(a million/sinx) and u get..... a million/cosxsinx=a million/cosxsinx =]

From Anonymous: left part
tan x = sinx / cosx
cot x = cosx...
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
Sat, 09 Jan 2010 11:57:57 +0000
left part
tan x = sinx / cosx
cot x = cosx / sinx
sin²x + cos²x = 1
((sinx)² + (cosx)²) / (cosx*sinx)
= 1 / (cosx* sinx)
right part
sec = 1 / cosx
csec = 1 / sinx
1 / (sinx*cosx)
so left = right

From Parker: Write everything in terms of sin and cos:
(...
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
https://answers.yahoo.com/question/index?qid=20100109114934AAGEucG
Sat, 09 Jan 2010 11:56:24 +0000
Write everything in terms of sin and cos:
(sinx/cosx)+(cosx/sinx)=(1/cosx)(1/sinx)
Find a common denominator of cosx*sinx on the left side and simplify the right side:
(sin^2x+cos^2x)/(sinx*cosx)=1/(sinx*cosx)
Realize that the numerator of the left side is an identity that equals 1. Thus:
1/(sinx*cosx)=1/(sinx*cosx)