Schaum's Outline of Theory and Problems of Beginning Calculus

216 THE TANGENT AND OTHER TRIGONOMETRIC FUNCTIONS [CHAP. 28Tkorem 283 (Identities): tan’ x + 1 = sec’ x and cot’ x + 1 = csc’ x.Proof: Divide sin2 x + cos2 x = 1 by cos’ x or sin2 x.Traditional DefinitionsAs was the case with sin 8 and cos 8, the supplementary trigonometric functions were originallydefined only for an acute angle of a right triangle. Referring to Fig. 26-3, we haveopposite sidetan 8 =adjacent sideadjacent sidecot 8 =opposite sidehypo tenusesec 8 =adjacent sidehypotenusecsc e =opposite side28.1 Prove Theorem 28.2.Solved ProblemsDAtan x) = D,(-) sin xcos x(cos x)D,(sin x) - (sin x)D,(cos x)- [by the quotient rule](cos x)2(cos xxcos x) - (sin x)( -sin x)- [by Theorem 27.21cos2 xcos2 x + sin’ x 1- -[by Theorem 26.11cos2 x cos’ x= sec2 x-1D,(cot x) = D,((tan x)- l) = - DAtan x) [by the power chain rule](tan x )~-1 1 -1=--=(tan x )~(cos x )~(tan x cos x)’-1 sin x=- [tan x =(sin x)’I= -csc2 xDifferentiating the first identity of Theorem 28.3,2(tan x)(sec2 x) = 2(sec x)D,(sec x)and dividing through by 2(sec x), which is never zero, givesD,(sec x) = tan x sec xSimilarly, differentiation of the second identity of Theorem 28.3 givesD,(csc x) = -cot x csc x