Question

If `sec θ = 17/8` then prove that `(3 - 4 sin^2θ)/(4 cos^2θ - 3) = (3 - tan^2 θ)/(1 - 3 tan^2θ)`.

Answer 1

Given: `sec θ = 17/8` `("so" cos θ = 8/17, sin^2 θ = 225/289, tan^2 θ = 225/64)`.

To Prove: `(3 - 4 sin^2θ)/(4 cos^2θ - 3) = (3 - tan^2 θ)/(1 - 3 tan^2θ)`

Proof [Step-wise]:

1. Start with the left-hand side:

LHS = `(3 - 4 sin^2θ)/(4 cos^2θ - 3)`.

2. Use sin²θ + cos²θ = 1 to rewrite 3:

3 = 3(sin²θ + cos²θ)

3. Transform the numerator:

3 – 4 sin²θ = 3(sin²θ + cos²θ) – 4 sin²θ 

= 3 cos²θ – sin²θ

4. Transform the denominator:

4 cos²θ – 3 = 4 cos²θ – 3(sin²θ + cos²θ)

= cos²θ – 3 sin²θ

5. Therefore LHS = `(3 cos^2θ - sin^2θ)/(cos^2θ - 3 sin^2θ)`.

6. Now simplify the right-hand side:

RHS = `(3 - tan^2θ)/(1 - 3 tan^2θ)`

= `(3 - (sin^2θ/cos^2θ))/(1 - 3(sin^2θ/cos^2θ))`

Multiply numerator and denominator by `cos^2θ = (3 cos^2θ - sin^2θ)/(cos^2θ - 3 sin^2θ)`.

7. Hence LHS = RHS.

Optional numeric check using `sec θ = 17/8`:

`cos^2θ = 64/289`

`sin^2θ = 225/289`

`tan^2θ = 225/64` 

Substituting gives both sides = `33/611`, confirming the equality.

The identity `(3 - 4 sin^2θ)/(4 cos^2θ - 3) = (3 - tan^2 θ)/(1 - 3 tan^2θ)` is proved consistent with the given `sec θ = 17/8`.