use algebra to prove the identity?

Question

(cos x / 1-sin x) - tan x = (1/cos x)

Answer 1

(cos x/1-sin x)-tan x
=(cos x/1-sin x)-sin x/cos x
=(cos^2 x-sin x[1-sin x])/cos x(1-sinx) [LCD is cos x(1-sin x)]
(cos^2 x-sin x+sin^2 x)/cos x(1-sin x)
=(1-sin x)/cos x(1-sin x)[cos ^2x+sin^2 x=1]

=1/cos x [Proved]

Answer 2

LHS
cos x / (1 - sin x) - sin x / cos x
[ cos² x - sinx (1 - sinx) ] / (cos x (1 - sin x)
[ cos² x + sin² - sin x ] / (cos x (1 - sin x))
(1 - sin x ) / (cos x (1 - sin x))
1 / cos x

RHS
1 / cos x

LHS = RHS

Answer 3

(cos x /1-sin x ) - sin x/cos x = 1/cos x
cos x /1-sin x = (1+sin x)cos x
cos ^2x = (1+sin x) (1-sin x)
cos ^2x = 1 - sin^2x
cos^2x + sin^2x = 1 This is an idenity and you have proved your origional equation

Answer 4

in case you multiply the two factors via one million/one million + cos(x) then you somewhat get sin(x)/one million-cos^(2)x= one million/sin(x) through identity sin^(2)x + cos^(2)x = one million (examine as sin squared x plus cosine squared x equals one million) then you somewhat comprehend that one million - cos^(2)x is comparable to sin^(2)x, so once you replace that for the time of, you get sin(x)/sin^(2)x = sin(x), that's a genuine assertion because of the fact the former is reminiscent of x/x^(2), that's purely one million/x, that's reminiscent of the latter. wish this helps :) *notice: as quickly as I write sin^(2)x, the sin function is squared, no longer the x. sin^(2)x = sin(x) circumstances sin(x)