How to verify the identity 1-tan(x)/sin(x)=csc(x)-sec(x)

(1 - tanx) / sinx
= (1 - (sinx/cosx)) / sinx ............. using tan(x) = sin(x)/cos(x)
= (cosx - sinx) / (cosxsinx) .................... multiply each term by cos(x)
= [cosx / (cosxsinx)] - [sinx / (cosxsinx)] ................ write as two separate fractions
= (1/sinx) - (1/cosx) .......................... cancel were you can
= csc(x) - sec(x) ............. using 1/sin(x) = csc(x) and 1/cos(x) = sec(x)

Usually, in these cases, you want to make both sides look kinda the same so you should Rey to change the left side of the equation to the reciprocal identities like sec, csc, etc. Change tan(x) to sin(x)/cos(x) and change that to (1/csc(x))/(1/sec(x)) and then divide the two which would leave you with sec(x)/csc(x). Don't forget that is all over sin(x) so divide (1-(sec(x)/csc(x))) by (1/csc(x)) and then that would lead to multiplying the numerator of the previous fraction by csc(x)/1. Distribute the csc(x) to both the 1 and the fraction, simplify the fraction and then you will have both sides of the equation exactly the same.

(1-tanx)/sinx = cscx-secx
(1-(sinx/cosx))/sinx = cscx-secx
((cosx/cosx)-(sinx/cosx))/sinx = cscx-secx
((cosx-sinx)/cosx)/sinx = cscx-secx
((cosx-sinx)/cosx)/(sinx/1) = cscx-secx
((cosx-sinx)/cosx)(1/sinx) = cscx-secx
(cosx-sinx)/(sinxcosx) = cscx-secx
(cosx/(sinxcosx))-(sinx/(sinxcosx)) = cscx-secx
(1/sinx)-(1/cosx) = cscx-secx
cscx-secx = cscx-secx

You do the oppositee on each side :DDDD ur welcum