Question

Prove the following trigonometric identities.

`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`

Answer 1

We have prove that

`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`

We know that `sin^2 A + cos^2 A = 1`

So,

(2 + cot A + tan A)(sin A - cos A)

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

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

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

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

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

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

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

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

`= (-cos^3 A + sin^3 A)/(sin A cos A)`

`= (sin^3 A - cos^3 A)/(sin A cos A)`

`= (sin^2  A)/cos A - cos^2 A/sin A`

`= sin A/cos A sin A - cos A/sin A cos A`

`= tan A sin A -  cot A cos A`

= sin A tan A - cos A cot A

Now

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

`= sin^2 A/cos A - cos^2 A/sin A`

`= sin A sin A/cos a - cos A cos A/sin A`

= sin A tan A - cos A cot A

Hence proved.