take the messy part ...
csc^-1 (x/(x-1) ) represents an angle Ø so that
csc Ø = x/(x-1)
or
sinØ = (x-1)/x
so make a sketch of a right-angled triangle with hypotenuse x and opposite side x-1
call the adjacent side b
then b^2 + (x-1)^2 = x^2
b^2 + x^2 - 2x + 1 = x^2
b^2 = 2x - 1
b = √(2x-1) ----> the adjacent side
so we have to find tanØ
tanØ = opposite/adjacent
= (x-1)/√(2x - 1)
This approach works for most of these type of problems
I've done this one multiply times yet it never seems to work out, I'm supposed to simplify yet it never works out
tan(csc^-1(x/x-1))
my first step;
sin/cos(1/sin(x/x-1))
1 answer
1 answer