Question

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

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

Answer 1

See the proof below

Explanation:

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

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