Question

Prove this identity now!

Answer 1

ok (:

Answer 2

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

Answer 3

@imqwerty PLEASE HELLLP!

Answer 4

1st we look at LHS\[\frac{ 1-\cos \theta }{ 1+\cos \theta }\times \frac{ 1-\cos \theta }{ 1-\cos \theta}\]\[\frac{ (1-\cos \theta)^2 }{ 1-\cos^2 \theta }\]\[\frac{ (1-\cos \theta )^2}{ \sin^2\theta }\]

now we simplify the RHS\[\left( \frac{ 1 }{ \sin \theta } -\frac{ \cos \theta }{ \sin \theta}\right)\]\[\left( \frac{ 1-\cos \theta }{ \sin \theta } \right)^2 => \frac{ (1-\cos \theta)^2 }{ \sin^2 \theta}\]

LHS=RHS
hence proved :)

Answer 5

Wow, you're a genius! My head is hurting so much from trying to prove so many identities!!!

Answer 6

(: u just gotta make a good start
try to make similar structures like we did in the 1st step to get sin(theta) in the denominator

Answer 7

Thanks for the tip :D

Answer 8

np :D