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) ]

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