[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)

[s_cianci]

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]

(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]

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]

(cosx)(tanx+sinxcotx)=sinx+cos^2x

Left hand side =

cosx(sinx/cosx + sinx cosx /sinx)

= sinx + cos^2 x

= right hand side