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View Full Version : Pre calc identities


anderson17
May 3, 2010, 03:41 PM
How do prove this to be true?

(secθ/sinθ) - (sinθ/cosθ) = cotθ

ArcSine
May 4, 2010, 04:51 AM
Let's fiddle with the left-hand side. Re-write the first term as

\frac{1}{sin(x)cos(x)}

Multiply the second term by "1", in the form of \frac{sin(x)}{sin(x)} and it becomes \frac{sin^2(x)}{sin(x)cos(x)}

Now both terms of the LHS have the same denominator, and can therefore be combined. Next, apply the Pythagorean Identity to the resulting numerator, and take it from there...