I need help with the following problem:

[Guest]

I need help with this problem I tried everything I could think of but I can not get it. sec x/sin x - sin x/cos x = cot

 

[stapel]

I need help with this problem I tried everything I could think of but I can not get it. sec x/sin x - sin x/cos x = cot

Please reply with clarification. In particular, please include the argument of the cotangent on the right-hand side of the equation, and please provide the instructions.

When you reply, please include a clear listing of the steps you have tried thus far.

Thank you.

Eliz.

 

[Guest]

:shock: Well I am not sure what you mean, but it is similar to the equation you helped solve before. I basically have to verify the equation and prove that one side is equal to the other. I turned sin x/cos x into tan. I didn't see how that would help me but I continued and turned sec x into 1/cos. I again got completely lost and turned cot x into cos x/ sin x. I need help is the basic point and I would truly appreciate it if you helped me. Once again the problem is below:


(sec x/sin x) - (sin x/cos x) = cot x

 

[soroban]

Hello, Carlos!

I assume the problem is: \(\displaystyle \L\:\frac{\sec x}{\sin x}\,-\,\frac{\sin x}{\cos x} \;=\;\cot x\)

Did you subtract the two fractions on the left?

Get a common denominator: \(\displaystyle \L\frac{\sec x}{\sin x}\cdot\frac{\cos x}{\cos x}\,-\,\frac{\sin x}{\cos x}\cdot\frac{sin x}{\sin x}\)

Since \(\displaystyle (\sec x)(\cos x)\,=\,1\), we have: \(\displaystyle \L\:\frac{1 \,-\,\sin^2x}{\sin x\cdot\cos x}\;=\;\frac{\cos^2x}{\sin x\cdot\cos x}{\;=\;\frac{\cos x}{\sin x}\;=\; \cot x\)

 

[Guest]

Thanks!

I expected it to be much more complicated but I was obviously wrong. Thank both of you for you help I may be able to solve the other problems I had questions on now but if not I will ask you guys again. Thanks again