How do you prove this is an identity?

sec x sin x +cscx cos x = tan x + cot x

Verify the following identity:

sec(x) sin(x) + csc(x) cos(x) = tan(x) + cot(x)

Write cotangent as cosine/sine, cosecant as 1/sine, secant as 1/cosine and tangent as sine/cosine:

1/cos(x) sin(x) + 1/sin(x) cos(x) = ^?cos(x)/sin(x) + sin(x)/cos(x)

Put cos(x)/sin(x) + sin(x)/cos(x) over the common denominator sin(x) cos(x): cos(x)/sin(x) + sin(x)/cos(x) = (cos(x)^2 + sin(x)^2)/(cos(x) sin(x)):

(cos(x)^2 + sin(x)^2)/(cos(x) sin(x)) = ^?cos(x)/sin(x) + sin(x)/cos(x)

Put cos(x)/sin(x) + sin(x)/cos(x) over the common denominator sin(x) cos(x): cos(x)/sin(x) + sin(x)/cos(x) = (cos(x)^2 + sin(x)^2)/(cos(x) sin(x)):

((cos(x)^2 + sin(x)^2)/sin(x))/cos(x) = ^?(cos(x)^2 + sin(x)^2)/(cos(x) sin(x))

The left hand side and right hand side are identical:

(identity has been verified)

Step-by-step explanation:

109 trust me

step-by-step explanation:

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.