Question

Fill in the blank !! question in picture below i have no clue what to do

Answer 1

(tanx)^2

Proof:

Verify the following identity:
(sec(x) - 1) (sec(x) + 1) = tan(x)^2

Write secant as 1/cosine and tangent as sine/cosine:
(1/cos(x) - 1) (1/cos(x) + 1) = ^? (sin(x)/cos(x) )^2

(sin(x)/cos(x))^2 = (sin(x)^2)/(cos(x)^2):
(1/cos(x) - 1) (1/cos(x) + 1) = ^?(sin(x)^2)/(cos(x)^2)

Put 1/cos(x) - 1 over the common denominator cos(x): 1/cos(x) - 1 = (1 - cos(x))/cos(x):
(1 - cos(x))/cos(x) (1/cos(x) + 1) = ^?(sin(x)^2)/(cos(x)^2)

Put 1/cos(x) + 1 over the common denominator cos(x): 1/cos(x) + 1 = (cos(x) + 1)/cos(x):
(1 - cos(x))/cos(x) (cos(x) + 1)/cos(x) = ^?(sin(x)^2)/(cos(x)^2)

((1 - cos(x)) (cos(x) + 1))/(cos(x) cos(x)) = ((1 - cos(x)) (cos(x) + 1))/cos(x)^2:
((1 - cos(x)) (cos(x) + 1))/(cos(x)^2) = ^?(sin(x)^2)/(cos(x)^2)

Multiply both sides by cos(x)^2:
(1 - cos(x)) (cos(x) + 1) = ^?sin(x)^2

(1 - cos(x)) (cos(x) + 1) = 1 - cos(x)^2:
1 - cos(x)^2 = ^?sin(x)^2

sin(x)^2 = 1 - cos(x)^2:
1 - cos(x)^2 = ^?1 - cos(x)^2

The left-hand side and right-hand side are identical:
Answer: (identity has been verified)