Question

I have to verify that this is true using trig identities.. Let me type it real quick

Answer 1

\[((\sin \theta)/(1-\cos \theta))-(\sin \theta \cos \theta)/(1+\cos \theta)=\csc \theta(1+\cos ^2\theta)\]

Answer 2

take common denominator
u can solve it easily

Answer 3

\[\frac{(\sin \theta)(1+\cos \theta)}{(1-\cos \theta)(1+\cos \theta)}-\frac{(\sin \theta \cos \theta)(1-\cos \theta)}{(1-\cos \theta)(1+\cos \theta)}\]

Answer 4

let teta be U
[sin U(1+cosU)-(1-cosU)(sinUcosU)]/1-cos^2 U
=[sinU+cosUsinU-sinUcosU+sinUcos^2U]/sin^2 U
=csc U+cscUcos^2U
=cscU(1+cos^2U)

Answer 5

\[\frac{1+\cos \theta}{\sin \theta}-\frac{\cos \theta(1-\cos \theta)}{\sin \theta}=\frac{1+\cos \theta-\cos \theta+\cos ^{2}\theta}{\sin \theta}\]\[\frac{(\sin \theta)(1+\cos \theta)}{\sin ^{2}\theta}-\frac{(\sin \theta \cos \theta)(1-\cos \theta)}{\sin ^{2}\theta}\]

Answer 6

\[\frac{1+\cos \theta}{\sin \theta}-\frac{(\cos \theta)(1-\cos \theta)}{\sin \theta}=\frac{1+\cos \theta-\cos \theta+\cos ^{2}\theta}{\sin \theta}\]

Answer 7

\[\csc \theta(1+\cos ^{2}\theta)\]