Could someone please help me prove this identity? #1/(secA-1) + 1/(secA+1) = 2cotAcosecA#

Question

#1/(secA-1) + 1/(secA+1) = 2cotAcosecA#

2574 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-08T07:56:57

See the proof below

We need

#1+tan^2A=sec^2A#

#secA=1/cosA#

#cotA=cosA/sinA#

#cscA=1/sinA#

Therefore,

#LHS=1/(secA+1)+1/(secA-1)#

#=(secA-1+secA+1)/((seca+1)(secA-1))#

#=(2secA)/(sec^2A-1)#

#=(2secA)/(tan^2A)#

#=2secA/(sin^2A/cos^2A)#

#=2/cosA*cos^2A/sin^2A#

#=2*cosA/sinA*1/sinA#

#=2cotAcscA#

#=RHS#

#QED#

Answer 2 · 2018-02-08T08:08:50

Please recall that

#sec A = 1 / (cos A)#

#1/(1/cos A -1) + 1/(1/cos A+1#

#cos A/(1-cos A)+cos A/(1+cosA)#

#(cos A+cos^2A+cosA-cos^2A)/(1-cos^2A)#

#(2 cosA )/(1-cos^2A)#

As #sin^2A+cos^2= 1# , we can rewrite the denominator like the following

#(2cosA)/sin^2A#

#(2cosA)/sinA 1/sin A#

Please remember that #cosA/sinA = cot A# and #1/sinA = cosecA#

Thus this leaves us with

#2cotA cosecA#

I hope this was helpful