Trigonometry (Identities)?

Question

Proove the following trigonometric equation using the trig identities (Pythagorean, Reciprocal, and Ratio).?
You need to use either the pythagorean, reciprocal, or ratio method for prooving this equation using identities. You can combine any of these three.

(1+cot / cotcsc) - (1-cot/csc) = sec

Show every step to get the best answer

Answer 1

Would it be sick and twisted if I said I had fun solving this? Anyway, here you go, hope you can read it... http://i3.photobucket.com/albums/y93/oomps62/trig.jpg

Answer 2

LHS = Tan^2 a / (a million + tan^2 a) Tan = sin/cos, so =[Sin^2(a)/Cos^2(a)]/[a million+Sin^2(a)/cos^2... placed the expression on the backside over a user-friendly denominator = [sin^2(a)/cos^2(a)]/[(cos^2(a)+sin^2(a))... sin^2 + cos^2 = a million, so = [sin^2(a)/cos^2(a)]/[a million/cos^2(a)] = sin^2(a)/cos^2(a) * cos^2(a)/a million = sin^2(a) = RHS

Answer 3

Prove the identity.
(1 + cot / cot*csc) - (1 - cot / csc) = sec

I think you meant:
(1 + cot) / (cot*csc) - (1 - cot) / csc = sec

Watch the parentheses.

Let's work with the left hand side.

Left Hand Side = (1 + cot) / (cot*csc) - (1 - cot) / csc

= (sin²)(1 + cot) / (cot*csc*sin²) - (sin)(1 - cot) / (csc*sin)

= (sin² + sin*cos)/cos - (sin - cos)/1

= (sin² + sin*cos)/cos - cos(sin - cos)/cos

= (sin² + sin*cos - sin*cos + cos²)/cos

= 1/cos = sec = Right Hand Side
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