Verifying Trig Identities: csc(A-B)=secB

Click For Summary
SUMMARY

The discussion centers on verifying the trigonometric identity \( \csc(A-B) = \frac{\sec B}{\sin A - \cos A \tan B} \). Participants emphasize the importance of rewriting the left-hand side in terms of \( \sin(A-B) \) rather than incorrectly distributing \( \csc \). The correct approach involves recognizing that \( \csc(A-B) \) equals \( \frac{1}{\sin(A-B)} \) and applying the sine subtraction formula. This method leads to a clearer path for verification.

PREREQUISITES
  • Understanding of trigonometric identities, specifically sine and cosecant functions.
  • Familiarity with the sine subtraction formula: \( \sin(A-B) = \sin A \cos B - \cos A \sin B \).
  • Knowledge of secant and tangent functions and their relationships to sine and cosine.
  • Ability to manipulate algebraic expressions involving trigonometric functions.
NEXT STEPS
  • Study the sine subtraction formula in detail to understand its application in trigonometric identities.
  • Learn how to convert between different trigonometric functions, particularly how to express secant and cosecant in terms of sine and cosine.
  • Practice verifying other trigonometric identities using similar methods of rewriting and substitution.
  • Explore common mistakes in trigonometric identity verification to avoid pitfalls in future problems.
USEFUL FOR

Students studying trigonometry, educators teaching trigonometric identities, and anyone seeking to improve their problem-solving skills in mathematics.

gunnar14
Messages
1
Reaction score
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 canceled 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 I am stuck here... help please
 
Physics news on Phys.org
gunnar14 said:

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 canceled 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 I am stuck here... help please
Just to be clear, the problem you're trying to solve is to verify the identity

\csc(A-B)=\frac{\sec B}{\sin A-\cos A \tan B}

right? Your post is kind of hard to read.

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

Similar threads

Replies
25
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
54
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K