Show that :

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

Solving, L.H.S. of the equation, we get :

$\Rightarrow \dfrac{\text{cos}^2 (45° + θ) + \text{cos}^2 (45° - θ)}{\text{tan (60° + θ) \text{tan (30° - θ)}}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 (45° + θ) + \text{sin}^2 [90° - (45° - θ)]}{\text{tan (60° + θ) \text{cot [90° - (30° - θ)]}}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 (45° + θ) + \text{sin}^2 (45° + θ)}{\text{tan (60° + θ) \text{cot (60° + θ)}}} \\[1em] \text{As, cos}^2 \text{ A + sin}^2 A = 1 \text{ and tan A. cot A} = 1 \\[1em] \Rightarrow \dfrac{1}{1} \\[1em] \Rightarrow 1.$

Since, L.H.S. = R.H.S.

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