Question

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

Answer 1

TRUE,`" " " "sin x/csc x+cos x/sec x=sin x csc x`

Explanation:

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 ...

Answer 2

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 !!