Question

If `(sec θ - tan θ) = sqrt(2) tan θ` then prove that `(sec θ + tan θ) = sqrt(2) sec θ`.

Answer 1

Given: `(sec θ - tan θ) = sqrt(2) · tan θ`.

To Prove: `(sec θ + tan θ) = sqrt(2) · sec θ`.

Proof [Step-wise]:

1. Start from the given:

`sec θ - tan θ = sqrt(2) · tan θ`

2. Solve for sec θ in terms of tan θ:

`sec θ = tan θ + sqrt(2) · tan θ`

= `(1 + sqrt(2)) · tan θ`   ...(1)

3. Express tan θ in terms of sec θ using (1):

`tan θ = (sec θ)/(1 + sqrt(2))`   ...(2)

4. Compute sec θ + tan θ using (2):

`sec θ + tan θ = sec θ + (sec θ)/(1 + sqrt(2))` 

= `sec θ (1 + 1/(1 + sqrt(2)))`

= `sec θ ((1 + sqrt(2) + 1)/(1 + sqrt(2)))` 

= `sec θ ((2 + sqrt(2))/(1 + sqrt(2)))`   ...(3)

5. Simplify the factor `(2 + sqrt(2))/(1 + sqrt(2))`:

`(2 + sqrt(2)) = sqrt(2)(1 + sqrt(2))` because `sqrt(2)(1 + sqrt(2)) = sqrt(2) + 2 = 2 + sqrt(2)`.

Therefore `(2 + sqrt(2))/(1 + sqrt(2)) = sqrt(2)`.

6. Substitute this into (3):

`sec θ + tan θ = sec θ · sqrt(2)`

= `sqrt(2) · sec θ`

`(sec θ + tan θ) = sqrt(2) · sec θ`, as required.