Show that $\frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\cos ^{2}\left(45^{\circ}-\theta\right)}{\tan \left(60^{\circ}+\theta\right) \tan \left(30^{\circ}-\theta\right)}=1$

(N/A) $L$.$H$.$S$. $= \frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\cos ^{2}\left(45^{\circ}-\theta\right)}{\tan \left(60^{\circ}+\theta\right) \cdot \tan \left(30^{\circ}-\theta\right)}$
Using the identity $\cos \theta = \sin(90^{\circ}-\theta)$ and $\tan \theta = \cot(90^{\circ}-\theta)$:
Numerator: $\cos ^{2}\left(45^{\circ}+\theta\right) + \sin ^{2}\left(90^{\circ}-(45^{\circ}-\theta)\right) = \cos ^{2}\left(45^{\circ}+\theta\right) + \sin ^{2}\left(45^{\circ}+\theta\right) = 1$
Denominator: $\tan \left(60^{\circ}+\theta\right) \cdot \cot \left(90^{\circ}-(30^{\circ}-\theta)\right) = \tan \left(60^{\circ}+\theta\right) \cdot \cot \left(60^{\circ}+\theta\right) = \tan \left(60^{\circ}+\theta\right) \cdot \frac{1}{\tan \left(60^{\circ}+\theta\right)} = 1$
Therefore,$\frac{1}{1} = 1 = \text{R.H.S.}$