Tan inverse cos x/1-sin x
write in simplest form?

Tan [arccos(x) ] / [1 - sin(x) ]

1) First work the numerator, which is the confussing part

x = cos(y) =>  arccos(x) = y

tan [arccos(x) ] = tan (y) = sin (y) / cos (y)

using the fundamental identity

[cos(y)] ^2 + [sin(y)}^2 = 1 => sin (y) = √(1 - [cos(y)]^2 ) = √(1 - x^2)

=> tan (y) = sin(y) / cos(y) = √(1-x^2) / x  ... this is the numerator

2) write the complete fraction using the numerator just found

[√(1 - x^2) / x] / (1 - sin(x) ) = √(1 - x^2) / [x (1 - sin(x) ]

√(1 - x^2) / [x (1 - sin(x) ]