15) will mark BRAINLIEST AND 10 POINTS.

Mathematics

2 answers:

vaieri [72.5K]2 years ago

6 0

Equation identically true

Ksivusya [100]2 years ago

6 0

Cotx=cosx/sinx
cot²x=cos²x/sin²x
cscx=1/sinx, cscx+1=1/sinx +sinx/sinx=(1+sinx)/sinx

cot²x/(cscx+1)=(cos²x/sin²x)*(sinx/(1+sinx)=cos²x/[sinx(1+sinx)]

Recall that sin²x+cos²x=1, so cos²x=1-sin²x=(1+sinx)(1-sinx)

therefore: cos²x/[sinx(1+sinx)] can be written as
(1+sinx)(1-sinx)//[sinx(1+sinx)]=(1-sinx)/sinx=the left side.

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