Prove? cscx(1+cosx)(cscx-cotx)=1cscx(1+cosx)(cscx−cotx)=1

3 Answers
sente
Dec 30, 2016

See explanation

Explanation:

We will use the following:

  • (a+b)(a-b) = a^2-b^2(a+b)(a−b)=a2−b2
  • csc(x) = 1/sin(x)csc(x)=1sin(x)
  • cot(x) = cos(x)/sin(x)cot(x)=cos(x)sin(x)
  • 1-cos^2(x) = sin^2(x)1−cos2(x)=sin2(x)

With those,

csc(x)(1+cos(x))(csc(x)-cot(x))csc(x)(1+cos(x))(csc(x)−cot(x))

= (csc(x)+cot(x))(csc(x)-cot(x))=(csc(x)+cot(x))(csc(x)−cot(x))

=csc^2(x)-cot^2(x)=csc2(x)−cot2(x)

=1/sin^2(x)-cos^2(x)/sin^2(x)=1sin2(x)−cos2(x)sin2(x)

=(1-cos^2(x))/sin^2(x)=1−cos2(x)sin2(x)

=sin^2(x)/sin^2(x)=sin2(x)sin2(x)

=1=1

P dilip_k
Dec 30, 2016

LHS=cosectheta(1+costheta)(cosectheta-cottheta)LHS=cosecθ(1+cosθ)(cosecθ−cotθ)

=(cosectheta+cosecthetacostheta)(cosectheta-cottheta)=(cosecθ+cosecθcosθ)(cosecθ−cotθ)

=(cosectheta+1/sinthetaxxcostheta)(cosectheta-cottheta)=(cosecθ+1sinθ×cosθ)(cosecθ−cotθ)

=(cosectheta+cottheta)(cosectheta-cottheta)=(cosecθ+cotθ)(cosecθ−cotθ)

=(cosec^2theta-cot^2theta)=1=RHS=(cosec2θ−cot2θ)=1=RHS

proved

MathFact-orials.blogspot.com
Dec 30, 2016

See below:

Explanation:

We have:

cscx(1+cosx)(cscx-cotx)=1cscx(1+cosx)(cscx−cotx)=1

distributing out:

(cscx+cscxcosx)(cscx-cotx)=1(cscx+cscxcosx)(cscx−cotx)=1

csc^2x-cscxcotx+csc^2xcosx-cscxcosxcotx=1csc2x−cscxcotx+csc2xcosx−cscxcosxcotx=1

and now to reorder and simplify:

1/sin^2x-cancel(cosx/sin^2x)+cancel(cosx/sin^2x)-cos^2x/sin^2x=1

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

Now we'll use the identity of sin^2x+cos^2x=1 and so sin^2x=1-cos^2x

(sin^2x)/sin^2x=1

1=1

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