How do you prove #sinx/cscx + cosx/secx = sinx cscx#?

Take note first that #sin x csc x = 1#

Why? because, #sin x=1/csc x#

multiplying both sides by csc x

#sin x * csc x= 1/csc x*csc x#

#sin x*csc x=1/cancelcsc x*cancelcsc x#

and

#sin x csc x = 1#

Take note also that

#sec x cos x=1# because #sec x=1/cos x#

Let us go back to the problem

#sin x/csc x+cos x/sec x=sin x csc x#

#sin x/csc x*sin x/sin x+cos x/sec x*cos x/cos x=sin x csc x#

because #sin x/sin x=1# and #cos x/cos x=1#

You can always multiply any quantity by 1 and will not change anything

#sin x/csc x*sin x/sin x+cos x/sec x*cos x/cos x=sin x csc x#

Next

#sin^2 x/(csc x sin x)+cos^2 x/(sec x cos x)=sin x csc x#

#sin^2 x/1+cos^2 x/1=sin x csc x#

#sin^2 x+cos^2 x=sin x csc x#

Also from Pythagorean Relation,

#sin^2 x+cos^2 x=1#

therefore

#sin^2 x+cos^2 x=sin x csc x#

becomes

#1=sin x csc x#

#sin x csc x=sin x csc x# TRUE !!!

God bless America ...

Another way is

#sin x/csc x+cos x/sec x=sin x csc x#

Starting from left, factor out #sin x#

#sin x(1/csc x+cos x/(sin x sec x))=sin x csc x#

#sin x(1/csc x+cos x/sin x*1/sec x)=sin x csc x#

From reciprocal relations: #1/csc x=sin x# and #1/sec x=cos x#
it follows

#sin x(sin x+cos x/sin x*cos x)=sin x csc x#

and then

#sin x(sin x+cos^2 x/sin x)=sin x csc x#

Simplify by using #sin x# as Least Common Denominator

#sin x(sin^2 x/sin x+cos^2 x/sin x)=sin x csc x#

#sin x((sin^2 x+cos^2 x)/sin x)=sin x csc x#

#sin x(1/sin x)=sin x csc x#

#sin x csc x=sin x csc x#

Proven !!