1. ## verify 1/secx-tanx= tanx+secx

I don't even know where to start with this, anytime I think I'm getting somewhere I hit a wall. can someone help me?

2. Originally Posted by andie38290
I don't even know where to start with this, anytime I think I'm getting somewhere I hit a wall. can someone help me?
We are good at helping people when they hit a wall - BUT if we don't see your work we don't know what the wall is!

In general, my advice on proving trig identities is
1) change all functions on one side to sine and cosine
2) use the fundamental identity, cos^2(x) + sin^2(x) = 1
3) if necessary, reconvert from sine and cosine to the function on the other side

3. 1/(1/cosx)-(sinx/cosx)=(sinx/cosx)+(1/cosx)

1/(1-sinx/cosx)=sinx+1/cosx

and then i get stuck.

4. ## verify 1/(secx-tanx)= tanx+secx

Originally Posted by andie38290
verify 1/(secx - tanx) = tanx + secx

1/[(1/cosx) - (sinx/cosx)] = (sinx/cosx) + (1/cosx)

. . .

and then i get stuck.
Need extra parentheses when typing inline - must make order-of-operations correct.

In the olden days we were only allowed to manipulate one side of the identity - but it is useful to know (by looking at the right sdde) that you are going to want the denominator to be cosx. In the meantime, multiply the left side by cosx/cosx:

L = cosx/(1 - sinx)

Now it is almost always useful to utilize cos^2x + sin^2x = 1. I can see if we multiply both the numerator and the denominator by (1 + sinx), the denominator will be the product of sum and difference of same two numbers (that is like "rationalizing" the denominator):

L = [cosx (1 + sinx)] / [(1 - sinx)(1 + sinx)]

YES! I can see now that it is going to work out. Can you finish?

5. Another way:

Left-hand-side:

$\dfrac{1}{sec(x) - tan(x)}$

$= \ \dfrac{1}{sec(x) - tan(x)} * \dfrac{sec(x) + tan(x)}{sec(x) + tan(x)}$

$= \ \dfrac{sec(x) + tan(x)}{sec^2(x) - tan^2(x)}$

Now use

$1 + tan^2(x) \ = \ sec^2(x)$

and you should be done.....

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