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

Question

2745 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-30T10:14:30

See explanation

We will use the following:

With those,

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

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

#=csc^2(x)-cot^2(x)#

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

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

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

#=1#

Answer 2 · 2016-12-30T10:18:07

#LHS=cosectheta(1+costheta)(cosectheta-cottheta)#

#=(cosectheta+cosecthetacostheta)(cosectheta-cottheta)#

#=(cosectheta+1/sinthetaxxcostheta)(cosectheta-cottheta)#

#=(cosectheta+cottheta)(cosectheta-cottheta)#

#=(cosec^2theta-cot^2theta)=1=RHS#

proved

Answer 3 · 2016-12-30T10:22:10

See below:

We have:

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

distributing out:

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

#csc^2x-cscxcotx+csc^2xcosx-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#