sec2(x) - tan2(x)= 1/cos2(x) - sin2(x)/cos2(x)= (1 - sin2(x)) / cos2(x)= cos2(x) / cos2(x)= 1
Let s = sin x; c = cos x. By definition, sec x = 1/cos x = 1/c; and tan x = (sin x) / (cos x) = s/c. We know, also, that s2 + c2 = 1. Then, dividing through by c2, we have, (s2/c2) + 1 = (1/c2), or (s/c)2 + 1 = (1/c)2; in other words, we have, tan2 x + 1 = sec2 x.
tan2 x + 1 = sec2 x ⇒ 1 - sec2 x = -tan2 x ⇒ (tan2 x - 1) / (1 - sec2 x) = (tan2 x - 1) / -tan2 x = (tan2 x) / (-tan2 x) - 1 / (-tan2 x) = -1 + cot2 x = cot2 x - 1 If you cannot remember tan2 x + 1 = sec2 x, remember and start from: sin2 x + cos2 x = 1 (which is used often) and divide each side by cos2 x: (sin2 x + cos2 x) / cos2 x = 1 / cos2 x ⇒ sin2 x / cos2 x + cos2 x / cos2 x = 1 / cos2 x But sin x / cos x = tan x; and 1/cos x = sec x ⇒ tan2 x + 1 = sec2 x Also, cot x = 1/tan x
cos2 x + sin2 x = 1 cos2 x = 1 - sin2 x
Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
sec2(x) - tan2(x)= 1/cos2(x) - sin2(x)/cos2(x)= (1 - sin2(x)) / cos2(x)= cos2(x) / cos2(x)= 1
Use these identities: sin2(x) + cos2(x) = 1, and tan(x) = sin(x)/cos(x) For clarity, the functions are written here without their arguments (the "of x" part). (1 - sin2) = cos2 (1 + tan2) = (1 + sin2/cos2) = (cos2+sin2) / cos2 = 1/cos2 Multiply them: (cos2) times (1/cos2) = 1'QED'
-cos2(x)1 = sin2(x) +cos2(x)1 - cos2(x) = sin2(x)-cos2(x) = sin2(x) - 1
3
Let s = sin x; c = cos x. By definition, sec x = 1/cos x = 1/c; and tan x = (sin x) / (cos x) = s/c. We know, also, that s2 + c2 = 1. Then, dividing through by c2, we have, (s2/c2) + 1 = (1/c2), or (s/c)2 + 1 = (1/c)2; in other words, we have, tan2 x + 1 = sec2 x.
Sin2(x)/Cos2(x) is an expression, not an equation. Because it is an expression, it cannot be solved. It can be transformed to other, equivalent expressions, but that is as far as you can go. So, Sin2(x)/Cos2(x) = [Sin(x)/Cos(x)]2 = Tan2x or [1/Cos2(x) - 1] or [Sec2(x) - 1]
tan2 x + 1 = sec2 x ⇒ 1 - sec2 x = -tan2 x ⇒ (tan2 x - 1) / (1 - sec2 x) = (tan2 x - 1) / -tan2 x = (tan2 x) / (-tan2 x) - 1 / (-tan2 x) = -1 + cot2 x = cot2 x - 1 If you cannot remember tan2 x + 1 = sec2 x, remember and start from: sin2 x + cos2 x = 1 (which is used often) and divide each side by cos2 x: (sin2 x + cos2 x) / cos2 x = 1 / cos2 x ⇒ sin2 x / cos2 x + cos2 x / cos2 x = 1 / cos2 x But sin x / cos x = tan x; and 1/cos x = sec x ⇒ tan2 x + 1 = sec2 x Also, cot x = 1/tan x
cos2 x + sin2 x = 1 cos2 x = 1 - sin2 x
x=1
sin x times sin x. or 1/cosec2(x) or 1 - cos2(x) or tan2(x)*cos2(x) etc, etc.
Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
Cos(90 - x) = sin(x) so cos2(90 - x) = sin2(x)