- #1

- 4

- 0

## Homework Statement

cos+sintan/sinsec=csc

## Homework Equations

## The Attempt at a Solution

(cos+sin(sin/cos))/(sin/1/cos)

(cos+sin^2sincos)/(sin/cos)

(cos+sin^2sincos)X(cos/sin)

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter fouracres
- Start date

- #1

- 4

- 0

cos+sintan/sinsec=csc

(cos+sin(sin/cos))/(sin/1/cos)

(cos+sin^2sincos)/(sin/cos)

(cos+sin^2sincos)X(cos/sin)

- #2

- 590

- 0

(cos+sin(sin/cos))/(sin/1/cos) to

(cos+sin^2sincos)/(sin/cos) ?

The cosine should become the LCM in the numerator.

- #3

- 4

- 0

- #4

- 4

- 0

where were u saying cos shud become the LCM in the numerator?

- #5

Mark44

Mentor

- 36,063

- 7,997

Have you ever heard of punctuation? The quote above appears to be three different sentences. Using periods at the ends of your sentences would make what you say easier to understand.

A similar problem exists with your first post (in addition to the omission of arguments of the functions shown):

cos+sintan/sinsec=csc

As you have written it, most people in this forum would interpret the above as:

[tex]cos(x) + \frac{sin(x)tan(x)}{sin(x)sec(x)} = csc(x)[/tex]

I suspect that what you really meant, though, was this:

[tex]\frac{cos(x) +sin(x)tan(x)}{sin(x)sec(x)} = csc(x)[/tex]

If you don't know how to use LaTeX, you can write the equation above like so:

(cos(x) +sin(x)tan(x))/(sin(x)sec(x))=csc(x)

Share: