Question

Prove the following trigonometric identities.

`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`

Prove the following:

`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`

Answer 1

In the given question, we need to prove `1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`

Here, we will first solve the L.H.S.

Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta`, we get

`1/(sec A +  tan A) - 1/cos A  = 1/(1/cos A + sin A/cos A) - (1/cos A)`

`= 1/(((1 + sin A)/cos A)) - (1/cos A)`

`= (cos A/(1 + sin A)) - (1/cos A)`

`= (cos^2 A - (1 + sin A))/((1 + sin A)(cos A))`

On further solving, we get

`(cos^2 A -(1 + sin A))/((1 + sin A)(cos A)) = (cos^2 A - 1 - sin A)/((1 +  sin A)(cos A))`

`= (-sin^2 A - sin A)/((1 + sin A)(cos A))`    (Using `sin^2 theta = 1 - cos^2 theta)`

`= (-sin A(sin A + 1))/((1 + sin A)(cos A))`

`= (-sin A)/cos A`

= − tan A

Similarly, we solve the R.H.S.

`((1 - sin A) - cos^2 A)/((cos A)(1 - sin^2 A)) = (1 - sin A - cos^2 A)/((cos A)(1 - sin A))`

`= (sin^2 A - sin A)/((cos A)(1 - sin A))`   (Using `sin^2 theta = 1- cos^2 theta`) 

`= (-sin A(1 - sin A))/((cos A)(1 - sin A))`

`= (-sin A)/cos A`

= − tan A

So, L.H.S = R.H.S

Hence proved.