sec(X)-tan(X)=cot(45+(X÷2))
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tan(x2+π4)=secx+tanx
Explanation:
Use the trig identity:
tan(a+b)=tana+tanb1−tana.tanb
Call t = (x/2) and develop the left side
LS=tan(t+π4)=tant+tan(π4)1−tant.tan(π4)
Trig table gives tan(π4)=1, there for:
LS=1+tant1−tant=(cost+sintcost)(costcost−sint)=
LS=cost+sintcost−sint
Multiply both numerator and denominator by (cos t + sin t), we get:
LS=(cost+sint)2cos2t−sin2t
Reminder:
(cost+sint)2=1+2cost.sint=1+sin2t
cos2t−sin2t=cos2t.
LS=1+sin2tcos2t=1cos2t+sin2tcos2t
LS=sec2t+tan2t
Replace t by (x2), we get
tan(x2+π4)=secx+tanx
Explanation:
Use the trig identity:
tan(a+b)=tana+tanb1−tana.tanb
Call t = (x/2) and develop the left side
LS=tan(t+π4)=tant+tan(π4)1−tant.tan(π4)
Trig table gives tan(π4)=1, there for:
LS=1+tant1−tant=(cost+sintcost)(costcost−sint)=
LS=cost+sintcost−sint
Multiply both numerator and denominator by (cos t + sin t), we get:
LS=(cost+sint)2cos2t−sin2t
Reminder:
(cost+sint)2=1+2cost.sint=1+sin2t
cos2t−sin2t=cos2t.
LS=1+sin2tcos2t=1cos2t+sin2tcos2t
LS=sec2t+tan2t
Replace t by (x2), we get
tan(x2+π4)=secx+tanx
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