Anonymous

# cosx(tanx+cotx)=cscx Prove?

How do I prove this?

Relevance
• kb
Lv 7

cos x (tan x + cot x)

= cos x [(sin x / cos x) + (cos x / sin x)]

= cos x [(sin^2(x) + cos^2(x)) / (sin x cos x)], common denominator

= [(sin^2(x) + cos^2(x)) / (sin x)]

= 1/ sin x

= csc x.

• Anonymous
5 years ago

multiply cosx with cotx. You get (cos^2)/sin. Make sin into (sin^2)/sin. Add them you get (cos^2+sin^2)/sin. Use your trig properties and you know that (cos^2+sin^2)=1 and 1/sin is csc.

• Anonymous

LeftHandSide = cosx(sinx/cosx + cosx/sinx)

= sinx + cosx^2/sinx

=sinx +(1-sinsx^2)/sinx

= (sinx^2 + 1 -sinx^2)/sinx

= 1/sinx = cscx = RightHandSide

QED

=cosx (sinx/cosx + cosx/sinx)

= sinx + (cosx)^2/ sinx

= (sinx)^2 / sinx + (cosx)^2/ sinx

= 1/sinx

= cscx

• TomV
Lv 7

cos(tan + cot) = csc

cos(sin/cos + cos/sin) = 1/sin

sincos(sin/cos + cos/sin) = 1

sin^2 + cos^2 = 1

1 = 1

cos(sin/cos + cos/sin) = 1/sin

sincos(sin/cos + cos/sin) = 1

sin^2 + cos^2 = 1

• Anonymous

CosX(TanX+CotX)=CscX

CosX(SinX/CosX+CosX/SinX)=CscX

SinxCosx/Cosx+CosxCosx/Sinx=CscX

SinX+(1-SinxSinx)/Sinx=CscX

SinX+1/sinx-Sinx=CscX

1/SinX=CscX

CscX=CscX