Question

If x = (sec A + sin A) and y = (sec A – sin A), prove that `(2/(x + y))^2 + ((x - y)/2)^2 = 1`.

Answer 1

Given:
x = sec A + sin A

y = sec A – sin A

To Prove: `(2/(x + y))^2 + ((x - y)/2)^2 = 1`

Proof (Step-wise):

1. Compute x + y:

x + y = (sec A + sin A) + (sec A – sin A)

= 2 sec A

2. Therefore `2/(x + y) = 2/(2 sec A)`

= `1/(sec A)`

= cos A

Hence `(2/(x + y))^2 = cos^2A`.

3. Compute x – y:

x – y = (sec A + sin A) – (sec A – sin A)

= 2 sin A

4. Therefore `(x - y)/2 = sin A`.

Hence `((x - y)/2)^2 = sin^2A`.

5. Add the two results:

`(2/(x + y))^2 + ((x - y)/2)^2 = cos^2 A + sin^2A`   ...(By Pythagorean identity)

= 1   

Thus `(2/(x + y))^2 + ((x - y)/2)^2 = 1`, as required.