Mastering Identities: Solving Tricky Problems in Pre-Cal 2

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SUMMARY

The discussion focuses on solving the identity problem Sec/Tan - Tan/Sec = cos(x)cot(x) in Pre-Calculus 2. The user expresses difficulty in selecting the correct identity and changing the denominator. The solution involves expanding sec and tan into their sine and cosine components and applying the arithmetic rule for dividing fractions. This method clarifies the steps needed to manipulate the expression effectively.

PREREQUISITES
  • Understanding of trigonometric identities, specifically secant and tangent.
  • Familiarity with fraction division and simplification techniques.
  • Basic knowledge of sine and cosine functions.
  • Ability to manipulate algebraic expressions involving trigonometric functions.
NEXT STEPS
  • Study the properties of trigonometric identities in depth.
  • Practice dividing and simplifying trigonometric fractions.
  • Explore the relationship between sine, cosine, secant, and tangent functions.
  • Learn advanced techniques for solving trigonometric equations.
USEFUL FOR

Students in Pre-Calculus, particularly those struggling with trigonometric identities and algebraic manipulation, as well as educators seeking to enhance their teaching methods in this area.

msdenise15
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Hi,

I'm new to this site and I'm very happy that I found it. My Pre-cal 2 teacher has been no help to me when it comes to explaining certain steps needed to solve a problem. Overall I'm having a hard time choosing the correct identity needed to solve the problems. However what I do not understand about this particular problem is how to change the denominator.

The question is:

Sec/Tan -Tan/Sec = cosxcotx

I started by expanding sec and tan which gave me (1/cos)/(sin/cos)-(cos/sin)/(1/cos)

but what should I do to change the denominator?
 
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That's a good start. Now think back to arithmetic: What do you do when you are asked to divide two fractions? I ask this, because that's exactly what you have there. For instance:

[tex]\frac{\frac{1}{\cos(x)}}{\frac{\sin(x)}{\cos(x)}}=\frac{1}{\cos(x)}\div\frac{\sin(x)}{\cos(x)}[/tex]

Does that help?
 

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