Yahoo Answers: Answers and Comments for Trig Identities Help Please!? [Physics]
Copyright © Yahoo! Inc. All rights reserved.
https://answers.yahoo.com/question/index?qid=20161205091437AAjqsKu
From Anonymous
enUS
Mon, 05 Dec 2016 09:14:38 +0000
3
Yahoo Answers: Answers and Comments for Trig Identities Help Please!? [Physics]
292
38
https://answers.yahoo.com/question/index?qid=20161205091437AAjqsKu
https://s.yimg.com/zz/combo?images/emaillogous.png

From NCS: Let me rewrite it as
sin(x)/(csc(x)1)  sin(...
https://answers.yahoo.com/question/index?qid=20161205091437AAjqsKu
https://answers.yahoo.com/question/index?qid=20161205091437AAjqsKu
Mon, 05 Dec 2016 13:00:06 +0000
Let me rewrite it as
sin(x)/(csc(x)1)  sin(x)/cot²(x) = tan²x
Dealing with the first term on the LHS:
csc(x) = 1/sin(x), so
LHS1 = sin(x)/(csc(x)  1) * sin(x)/sin(x) = sin²(x) / (1  sin(x))
... multiply by (1+sin(x)) / (1+sin(x))
LHS1 = sin²(x) / (1  sin(x)) * (1 + sin(x)) / (1 + sin(x))
LHS1 = sin²(x) * (1 + sin(x)) / (1  sin²(x))
... and 1  sin²(x) = cos²(x), so
LHS1 = sin²(x) * (1 + sin(x)) / cos²(x) = tan²(x) * (1 + sin(x))
Dealing with the second term on the LHS:
1/cot²(x) = tan²(x), so
LHS2 = tan²(x) * sin(x)
So
LHS = LHS1  LHS2 = tan²(x)*(1 + sin(x))  tan²(x)*sin(x)
LHS = tan²(x) * (1 + sin(x)  sin(x)) = tan²(x) = RHS
Hope this helps!