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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