Show that :

$\dfrac{\text{cos}^2 \text{(45° + θ)} + \text{cos}^2 \text{(45° - θ)}}{\text{tan(60° + θ) tan (30° - θ)}}$ = 1.

$\text{L.H.S.} = \dfrac{\text{cos}^2 \text{(45° + θ)} + \text{cos}^2 \text{(45° - θ)}}{\text{tan(60° + θ) tan (30° - θ)}}\\[1em] = \dfrac{\text{cos}^2 \text{((90° - 45°) + θ)} + \text{cos}^2 \text{(45° - θ)}}{\text{tan((90° - 30°) + θ) tan (30° - θ)}}\\[1em] = \dfrac{\text{cos}^2 \text{(90° - (45° - θ))} + \text{cos}^2 \text{(45° - θ)}}{\text{tan(90° - (30° - θ)) tan (30° - θ)}}\\[1em] = \dfrac{\text{sin}^2 \text{(45° - θ)} + \text{cos}^2 \text{(45° - θ)}}{\text{cot (30° - θ) tan (30° - θ)}}\\[1em] = \dfrac{1}{\dfrac{1}{\text{tan (30° - θ)}}{\text{tan (30° - θ)}}}\\[1em] = \dfrac{1}{\dfrac{1}{\cancel{tan (30° - θ)}}{\cancel{tan (30° - θ)}}}\\[1em] = 1$

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence, $\dfrac{\text{cos}^2 \text{(45° + θ)} + \text{cos}^2 \text{(45° - θ)}}{\text{tan(60° + θ) tan (30° - θ)}}$ = 1.