Question

Prove the following trigonometric identities.

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

Answer 1

We have to prove `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`

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

So,

`sin A/(sec A + tab A - 1) + cos A/(cosec A + cot A -1)`

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

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

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

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

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

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

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

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

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

`= (2 sin A cos A)/(2 sin A cos A)`

= 1

Hence proved.