Prove
sec2x - tan2x = tan(45-x)
Pls Help!!!TQ

Step-by-step explanation:

sec(2x) − tan(2x)

Write in terms of sine and cosine:

1/cos(2x) − sin(2x)/cos(2x)

(1 − sin(2x)) / cos(2x)

Use double angle formulas:

[ 1 − 2 tan x / (1 + tan²x) ] / [ (1 − tan²x) / (1 + tan²x) ]

Multiply by reciprocal:

[ 1 − 2 tan x / (1 + tan²x) ] × [ (1 + tan²x) / (1 − tan²x) ]

Distribute 1 + tan²x:

(1 + tan²x − 2 tan x) / (1 − tan²x)

Factor:

(1 − tan x)² / ((1 − tan x) (1 + tan x))

Simplify:

(1 − tan x) / (1 + tan x)

Rewrite using 1 = tan(45°):

(tan(45°) − tan x) / (tan(45°) + tan x)

Use angle difference formula:

tan(45° − x)

v = 1/3 a^2h

step-by-step explanation:

if the height is 57.7cm and the height of the slant edge is 18.2cm, then the volume is 1/3 a^2h.

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval