Question

Prove the following identities:

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

Answer 1

Given: `(tan A + sin A)/(tan A - sin A) = "?" (sec A + 1)/(sec A - 1)`

To Prove: `(tan A + sin A)/(tan A - sin A) = (sec A + 1)/(sec A - 1)`

Proof [Step-wise]:

1. Start with the left-hand side (LHS):

`LHS = (tan A + sin A)/(tan A - sin A)`

2. Write tan and sec in terms of sin and cos:

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

`sec A = 1/(cos A)`

3. Substitute `tan A = (sin A)/(cos A)` into LHS: 

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

4. Factor sin A from numerator and denominator:

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

5. Cancel the common factor sin A (provided sin A ≠ 0):

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

6. Replace `1/(cos A)` by sec A: 

LHS = `(sec A + 1)/(sec A - 1) = RHS`

Therefore `(tan A + sin A)/(tan A - sin A) = (sec A + 1)/(sec A - 1)`, for angles A where the expressions are defined in particular cos A ≠ 0, sin A ≠ 0, and sec A ≠ 1 so denominators are nonzero.