 Ask Remember Me? mathretard4592222 Posts: 1, Reputation: 1 New Member #1 Oct 22, 2007, 07:03 PM
Proofs? Identities?
Some one teach me precal I can't do any of this.

Prove the identity:
(cosx)(tanx+sinxcotx)=sinx+cos^2x (thats cos squared times x) s_cianci Posts: 5,472, Reputation: 760 Uber Member #2 Oct 22, 2007, 07:05 PM
Change everything to sin and cos. Recall that tan = sin/cos , cot = cos/sin , sec = 1/cos and csc = 1/sin. Take the left side of the equation, (cosx)(tanx+sinxcotx), do the algebra and it'll fall right into place. Kirandeep Singh Posts: 2, Reputation: 1 New Member #3 Oct 24, 2007, 07:13 AM
(cosx)(tanx + sinxcotx) = sinx + cos^2 x

Taking left hand side and putting tanx = sinx/cosx and cotx = cosx/sinx

cosx(sinx/cosx + sinx*cosx/sinx)
= sinx + cos^2x

Hence proved nycfunction Posts: 29, Reputation: 4 New Member #4 Nov 11, 2007, 12:05 PM Originally Posted by mathretard4592222
Some one teach me precal I can't do any of this.

Prove the identity:
(cosx)(tanx+sinxcotx)=sinx+cos^2x (thats cos squared times x)

tanx = sinx/cosx

cotx = cosx/sinx

On the left side:

(cosx)(tanx+sinxcotx)= cosx(sinx/cosx + sinx(cosx/sinx)

cosx(sinx/cosx) = sinx... cosx cancels out.

sinx(cosx/sinx) = cosx(cosx) = cos^2x

Final answer: sinx + cos^2(x) = right side terryg752 Posts: 197, Reputation: 4 Junior Member #5 Nov 11, 2007, 03:49 PM
(cosx)(tanx+sinxcotx)=sinx+cos^2x

Left hand side =

cosx(sinx/cosx + sinx cosx /sinx)

= sinx + cos^2 x

= right hand side

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