trigonometric identities
Hello,
(cotx - cscx)(cosx + 1) = - sinx
let's rewrite cotx as cosx /sinx and cscx as 1 /sinx:
[(cosx /sinx) - (1 /sinx)] (cosx + 1) = - sinx
[(cosx - 1) /sinx] (cosx + 1) = - sinx
[(cosx + 1)(cosx - 1)] /sinx = - sinx
(expanding the numerator)
(cos²x - 1) /sinx = - sinx
let's apply the identity 1 - cos²x = sin²x:
- (1 - cos²x) /sinx = - sinx
- sin²x /sinx = - sinx
ending with:
- sinx = - sinx (proved)
I hope it helps
(cosx/sinx-1/sinx)(cosx+1)
(cosx-1)(cosx+1)/sinx
(cos^2 x-1)/sinx
-sin^2 x/sinx
-sinx
Copyright © 2024 QUIZLS.COM - All rights reserved.
Answers & Comments
Verified answer
Hello,
(cotx - cscx)(cosx + 1) = - sinx
let's rewrite cotx as cosx /sinx and cscx as 1 /sinx:
[(cosx /sinx) - (1 /sinx)] (cosx + 1) = - sinx
[(cosx - 1) /sinx] (cosx + 1) = - sinx
[(cosx + 1)(cosx - 1)] /sinx = - sinx
(expanding the numerator)
(cos²x - 1) /sinx = - sinx
let's apply the identity 1 - cos²x = sin²x:
- (1 - cos²x) /sinx = - sinx
- sin²x /sinx = - sinx
ending with:
- sinx = - sinx (proved)
I hope it helps
(cosx/sinx-1/sinx)(cosx+1)
(cosx-1)(cosx+1)/sinx
(cos^2 x-1)/sinx
-sin^2 x/sinx
-sinx