Is this trigonometric solution correct?

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SUMMARY

The solution for the derivative y' from the equation 4[(cos x)(cos y)(y') + (sin y)(-sin x)] = 0 is confirmed as y' = (sin x sin y) / (cos x cos y). This expression can be further simplified using the identity tanA = sinA/cosA, leading to y' = tan x tan y. The discussion highlights the importance of recognizing trigonometric identities in simplifying expressions.

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Homework Statement



Solve for y':

4[(cos x)(cos y)(y') + (sin y)(-sin x)] = 0

Homework Equations



none

The Attempt at a Solution



4[(cos x)(cos y)(y') + (sin y)(-sin x)] = 0

4y' (cos x) (cos y) - 4 (sin x) (sin y) = 0

4y' (cos x) (cos y) = 4 (sin x) (sin y)

y' = [4 (sin x) (sin y)] / [4 (cos x) (cos y)]

y' = (sin x sin y) / (cos x cos y)


Thanks! I am really shaky at trigonometry, so the above worries me.
 
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Yes that is correct. If you want to, you can simplify it even more using the fact that sinA/cosA = tanA.
 
Thanks rock.freak667! See what I mean? I didn't even see that in the equation, although I do know that tan x = sin x/cos x.
 

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