I need a proof for the following trig identity

sinx + cosx = secx
cotx

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Originally Posted by brynscalltoarms

I need a proof for the following trig identity

sinx + cosx = secx
cotx

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Hint: on the left hand side of the equation convert the cot(x) function to its cos(x)/sin(x) equivalent, then factor out a 1/cos(x) term. What do you get?

Convert cot(x) into its alter ego in terms of sin and cos. Then play with the LHS of the equation to work it into a single fraction. See where that takes you.

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Originally Posted by ebaines

Hint: on the left hand side of the equation convert the cot(x) function to its cos(x)/sin(x) equivalent, then factor out a 1/cos(x) term. What do you get?

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But then I end up with sinx*cosx / sinx. In that case. I simplify to cosx + cosx. And cosx doesn't equal secx

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Originally Posted by brynscalltoarms

but then i end up with sinx*cosx / sinx. In that case. I simplify to cosx + cosx. and cosx doesn't equal secx

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There was supposed to be a two in front of that last cosx

?

Have another go at that part.

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Originally Posted by ArcSine

??

Have another go at that part.

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would that equal (sin^2x)/cosx?

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Originally Posted by ArcSine

??

Have another go at that part.

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oh no, that would turn into 1/cosx which is secx. But now i have this:

secx + cosx = secx

Quote:

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Originally Posted by brynscalltoarms

would that equal (sin^2x)/cosx?

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Yes. So now the left hand side is:

sin^2x/cosx + cosx

Now as I suggested earlier divide through by 1/cosx - what do you get?

Quote:

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Originally Posted by ebaines

Yes. So now the left hand side is:

sin^2x/cosx + cosx

Now as I suggested earlier divide through by 1/cosx - what do you get?

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I don't understand what you mean when you say to divide through by 1/cosx. I could multiply to get a common denominator of cosx. In which case my equation would be:
(sin^2x / cosx) + (cosx / cosx) [ or 1] = secx
So I could do:
((sin^2x + cosx) / cosx) = secx
to:
sin^2x = secx
onto:
1- cos^2x = secx
and so on from there...

Note that cos(x) =

Then combine the two LHS fractions over than common denominator.