# SOLUTION: verifying trigonometric identities. secx(tanx+cotx)=(cscx/cos^2x)

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 Question 592923: verifying trigonometric identities. secx(tanx+cotx)=(cscx/cos^2x)Found 2 solutions by Alan3354, jsmallt9:Answer by Alan3354(30983)   (Show Source): You can put this solution on YOUR website!verifying trigonometric identities. secx(tanx+cotx)=(cscx/cos^2x) -------------- Change to sines and cosines. Answer by jsmallt9(3296)   (Show Source): You can put this solution on YOUR website!These identities can be hard because there is no "recipe" that one can memorize and then apply to each one. When I looked at your problem no solution was immediately obvious to me. When this happens, I recommend that all sec's, csc's, tan's and cot's be replaced with equivalent expressions of sin and/or cos. doing this with your expression we get: Now we start using substitutions and/or Algebraic manipulations to transform the left side until we see a "path" to a solution. I'm going to start by using the Distributive Property to multiply out the left side. (This looks like a promising action because I can see that I will get a cos^2(x) in a denominator, just like the right side has!): Since the right side is just one term and the left side is two terms it looks like adding the fractions is promising. Of course we need to have the same denominators to add them: which gives us: Adding we get: The numerator is well-known: At this point I hope it is obvious how close to the finish we are. There are several ways to go from here. One technique which can be handy in many of these problems is to "manufacture" a missing part of the right side. We want a csc(x) in the numerator. We can "manufacture" one by: We get the numerator we want. And look, since csc and sin are reciprocals of each other, they cancel each other out when you multiply them in the denominator: And we're done!