How do you prove cotA+tan2A= cotAsec2A?

1 Answer
Gió · dani83
Aug 4, 2015

Probably there is a shorter and more elegant way but...

Explanation:

Considering that:
cot A=(cos A)/(sin A)
tan 2A=(sin2A)/(cos2A)=(2sinAcosA)/(cos^2A-sin^2A)
sec 2A=1/(cos2A)=1/(cos^2A-sin^2A)
sin^2 A + cos^2 A = 1

Using these identities into your expression you get:

cot A+tan 2A
= (cosA)/(sinA)+(2sinAcosA)/(cos^2A-sin^2A)
= (cosA(cos^2A-sin^2A)+2sin^2AcosA)/(sinA(cos^2A-sin^2A))
=(cosA(cos^2 A + sin^2 A))/(sinA(cos^2A-sin^2A))
= cot A sec 2A

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