How to Prove ??? #(sinA+cosA)/(cosA-SinA) = tan2A+sec2A#

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Answer 1 · 2018-05-21T03:27:14

#LHS=(sinA+cosA)/(cosA-SinA)#

#=(sinA+cosA)^2/((cosA-SinA)(cosA+sinA)#
#=(sin^2A+cos^2A+2sinAcosA)/(cos^2A-Sin^2A)#

#=(2sinAcosA+1)/(cos2A)#

#=(sin2A)/(cos2A)+1/(cos2A)#

# = tan2A+sec2A=RHS#

Answer 2 · 2018-05-21T03:33:44

It seemed easier to start from the right; please see proof below.

We need the crudest cosine double angle formula:

#cos(2A) = cos(A+A) = cos^2 A - sin ^2 A #

#sin(2A) = sin(A+A) = 2 sin A cos A #

#tan 2A + sec 2A #

#= {sin 2A}/{cos 2A} + 1/{cos 2A } #

# = {1 + sin 2A}/{cos 2A } #

# = { 1 + 2 sin A cos A}/{cos^2 A - sin ^2A } #

# ={ sin ^2 A + cos ^2 A + 2 sin A cos A }/{cos ^2 A- sin ^2 A}#

#= {(sin A + cos A)^2 }/{(cos A - sin A)(cos A+sin A)}#

#= {sin A + cos A}/{cos A - sin A} quad sqrt#