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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