Cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3 (x)
prove the trig identity

Working with the right side:

cot(x) + 2 tan(x) + tan³(x) = cos(x)/sin(x) + 2 sin(x)/cos(x) + sin³(x)/cos³(x)

… = (cos⁴(x) + 2 sin²(x) cos²(x) + sin⁴(x)) / (sin(x) cos³(x))

Factorize the numerator as a sum of squares:

a⁴ + 2 a² b² + b⁴ = (a² + b²)²

… = (cos²(x) + sin²(x))² / (sin(x) cos³(x))

Recall that

cos²(x) + sin²(x) = 1

… = 1 / (sin(x) cos³(x))

… = 1 / (sin(x) cos³(x)) • cos(x)/cos(x)

… = cos(x) / (sin(x) cos⁴(x))

… = cot(x) sec⁴(x)

the formula of a slope:

from a table we have

substitute:

the slope of f(x) is -1.

the graph of funcfion g(x) is a horizontal line. the slope of a horizontal line is equal 0. therefore the slope of g(x) is 0.

-1 < 0

step-by-step explanation:

the shape,the size and the distance