Mastering Identities: Solving Tricky Problems in Pre-Cal 2

[msdenise15]

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?

 

[quantumdude]

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:

$$\frac{\frac{1}{\cos(x)}}{\frac{\sin(x)}{\cos(x)}}=\frac{1}{\cos(x)}\div\frac{\sin(x)}{\cos(x)}$$

Does that help?

 

[Architeuthis Dux]

I understand the frustration of not being able to fully grasp a concept or solve a problem. Pre-calculus can be a challenging subject, but with the right approach, it can be mastered. I would suggest starting by reviewing the basic trigonometric identities, such as secant and tangent. These identities are crucial in solving trigonometric equations and understanding how to manipulate them can make solving problems much easier.

In this specific problem, you are on the right track by expanding secant and tangent. However, instead of trying to change the denominator, try simplifying the expression by factoring out a common term. In this case, both the numerator and denominator have a common factor of 1/cosx. By factoring this out, you will be left with 1/sinx - cosx/1, which can be simplified further using the reciprocal identity for sine and cosine.

I would also recommend practicing more problems and seeking help from your teacher or a tutor if needed. With perseverance and a solid understanding of the trigonometric identities, you will be able to master identities and solve tricky problems in pre-calculus. Best of luck!