Pre-calculus question?

Question

csc 45º x tan 60º

My process:

sin = √2 / 2

so csc = 1/ √2 / 2

times tan of 60º = √3/2 / 1/2

The answer is √6, but how?

Answer 1

csc45 is equal to inverse sin

sin45=√2 / 2

therefore csc45=2 / √2

tan60=√3 / 1

(2 / √2) (√3 / 1)

2√3 / √2

to get the radical out from the denominator

mulitply by the square root of two

√2 (2√3 / √2)

2√6 / 2

the two's cancel out and you're left with √6