2006-12-19 13:09:59 · 4 answers · asked by DadinTrouble 1 in Science & Mathematics ➔ Mathematics
= (cosx/sinx + sinx/cosx) * sin²x
= (sinxcos²x + sin³x) / cosx
= sinx (cos²x + sin²x) / cosx
= sinx/cosx
= tanx.
2006-12-19 13:16:25 · answer #1 · answered by Anonymous · 1⤊ 0⤋
(cot x + tan x) / csc ^ 2 x
changing csc ^ 2 x to 1/sin^2x you then get
(cot x + tan x) * sin^2x =
(cot x + tan x) * sin x * sin x =
(cot x * sin x * sin x) + (tan x * sin x * sin x)
(cos x * sin x) + (tan x * sin x * sin x)
2006-12-19 13:19:51 · answer #2 · answered by Renaud 3 · 0⤊ 0⤋
(cot(x) + tan(x))/(csc(x)^2)
((1/tan(x)) + tan(x))/((1/sin(x))^2)
((1 + tan(x)^2)/(tan(x)) / (1/sin(x)^2)
(sin(x)^2 * (1 + tan(x)^2))/(tan(x))
(sin(x)^2 * sec(x)^2)/(sin(x)/cos(x))
(sin(x)^2 * (1/cos(x)^2)) * (cos(x)/sin(x))
(sin(x)^2 * (cos(x)/cos(x)^2))/(sin(x))
sin(x) * (1/cos(x))
(sin(x)/cos(x))
tan(x)
ANS : tan(x)
2006-12-19 14:02:45 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
(cosx/sinx+sinx/cosx)*sin^2x
1/sinxcosx *sin^2x
=tanx
2006-12-19 13:14:39 · answer #4 · answered by raj 7 · 0⤊ 0⤋