How do you simplify the trigonometric expression (cos^2 x) (tan x + cot x)?

Answer 1

tan x = (sin x)/(cos x)
cot x = (cos x)/(sin x)
cos^2 x = 1 - sin^2 x

(cos^2 x) * (tan x + cot x)
= cos^2 x * (sin x)/(cos x) + (1- sin^2 x) * (cos x)/(sin x)
= (sin x) (cos x) + (cos x)/(sin x) - (sin x) (cos x)
= cot x

Answer 2

(cos ^2 x) (tan x + cot x)
(cos^2 x) ( sinx/cosx + cosx/sinx)
(cos^2 x) (sin^2 x + cos^2 x)/sinxcosx )
(cos^2 x) (1/sinxcosx)

cos^2 x * 1/sinxcosx
= cos x / sin x

= cot x

Answer 3

D is the respond you probably did it suited, you basically could desire to cancel between the cosx on perfect. cos^2x(tanx + cotx)= cos^2x (sinx/cosx + cosx/sinx) your easy denominator interior parentheses could be sinxcosx so cos^2x [(sin^2x + cos^2x)/sinxcosx) ; the needed id is that sin squared x + cos squared x = a million so cos^2x(a million/cosxsinx) = cosxcosx(a million/cosxsinx) certainly one of cosx on the perfect cancels out and all you're left with is cosx/sinx=cotx without the reason it is cos^2x(tanx + cotx)= cos^2x (sinx/cosx + cosx/sinx)= cos^2x [(sin^2x + cos^2x)/sinxcosx)= cos^2x(a million/cosxsinx)= cosxcosx(a million/cosxsinx)= cosx/sinx= cotx

Answer 4

your equation will simplify as

(cos^2 x)(sinx/cosx +cosx/sinx)
=(cos^2 x) [(sin^2 x +cos^2 x)/(sinx cosx)]
=cos x *1/sinx =cotx

Answer 5

cos ² x [(sin x / cos x) + (cos x / sin x) ]
sin x cos x + cos ³ x / sin x
cos x [ sin x + cos ² x / sin x ]
cos x [ (sin ² x + cos ² x) sin x ]
cos x [ 1 / sin x ]
cot x