Trig Identities

  • Thread starter gunnar14
  • Start date
  • #1
1
0

Homework Statement


Verify that each equation is an identity- directions
Problem- csc(A-B)=secB
---------------- <<< divide bar
sinA-cosAtanB

Homework Equations



well i tried to put in terms of sin cos and i've gotten stuck

The Attempt at a Solution



i started off by disributing i.e.>> cscA-cscB=secB/sinA-cosAtanB
next i changed tan to sin/cos and cancelled out cos so i was left with cscA-cscB=sec/sinA-sinB

changed csc to 1/sinA-1/sinB=sec/sinA-sinB

changed sec to 1/cos so i flipped and multiplied 1/sinA-sinB and got 1/sinA-1/sinB= 1/sincosA-sincosB and im stuck here..... help please
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,658
1,292

Homework Statement


Verify that each equation is an identity- directions
Problem- csc(A-B)=secB
---------------- <<< divide bar
sinA-cosAtanB

Homework Equations



well i tried to put in terms of sin cos and i've gotten stuck

The Attempt at a Solution



i started off by disributing i.e.>> cscA-cscB=secB/sinA-cosAtanB
next i changed tan to sin/cos and cancelled out cos so i was left with cscA-cscB=sec/sinA-sinB

changed csc to 1/sinA-1/sinB=sec/sinA-sinB

changed sec to 1/cos so i flipped and multiplied 1/sinA-sinB and got 1/sinA-1/sinB= 1/sincosA-sincosB and im stuck here..... help please
Just to be clear, the problem you're trying to solve is to verify the identity

[tex]\csc(A-B)=\frac{\sec B}{\sin A-\cos A \tan B}[/tex]

right? Your post is kind of hard to read.

Your very first step is wrong because [itex]\csc(A-B) \ne \csc A - \csc B[/itex]. Try writing it in terms of sin(A-B) first and go from there.
 

Related Threads on Trig Identities

  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
2
Views
779
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
1K
Top