Trigonometric identities: prove: (sin ß - cos ß - sin^3 ß)/sin ß = cos² ß - cot ß?

Answer 1

(sinA-cosA-sin^3A)/sinA
=1-cotA-sin^2A
=(1-sin^2A)-cotA
=cos^2A-cotA

Answer 2

(sin ß - cos ß - sin^3 ß)/sin ß = cos² ß - cot ß

LHS:
(sin ß - cos ß - sin^3 ß)/sin ß
As the expression staes, divide all 3 terms of the numerator by sin ß
=1-cot ß- sin² ß
=1-cot ß-(1-cos² ß)
=1-cot ß-1+cos² ß
=cos² ß- cot ß
=RHS

Note that the identity
sin² ß+cos² ß=1 should be known.
Thus, sin² ß= 1-cos² ß

Answer 3

(sin ß - cos ß - sin^3 ß)/sin ß = cos² ß - cot ß

Simplify the left side.

1 - (cos ß/sin ß) - sin^2 ß = cos² ß - cot ß

1 - cot ß - sin^2 ß = cos² ß - cot ß

Regroup terms.

(1 - sin^2 ß) - cot ß = cos² ß - cot ß

cos^2 ß - cot ß = cos² ß - cot ß

Answer 4

So simple!

(sin ß - cos ß - sin^3 ß)/sin ß
= sin ß / sin ß - cos ß / sin ß - sin^3 ß/sin ß
= 1 - cot ß - sin^2 ß
= ( 1 - sin^2 ß ) - cot ß
= cos^2 ß - cot ß

Rite?

Answer 5

(sin(ß) - cos(ß) - sin(ß)^3)/(sin(ß)) Divide each term.
1 - cot(ß) - sin²(ß) Now remember cos²(ß) + sin²(ß) = 1
so sin²(ß) = 1 - cos²(ß) and
1 - cot(ß) - (1 - cos²(ß)) is
cos²(ß) - cot(ß) QED

HTH ☺

Doug

Answer 6

(sin x - cos x - sin^3 x)/ sin x = [(sin x)(1-sin^2 x) - cos x] /sin x =

= 1 - sin^2 x - (cos x)/(sin x) = cos^2 x - cot x