If θ and 2θ − 45° are acute angles such that sin θ = cos (2θ − 45°),

Given that: $\sin \theta=\cos \left(2 \theta-45^{\circ}\right)$ and $\theta$ and $2 \theta-45$ are acute angles

We have to find $\tan \theta$

$\Rightarrow \sin \theta=\cos \left(2 \theta-45^{\circ}\right)$

$\Rightarrow \cos \left(90^{\circ}-\theta\right)=\cos \left(2 \theta-45^{\circ}\right)$

$\Rightarrow 90^{\circ}-\theta=2 \theta-45^{\circ}$

$\Rightarrow 3 \theta=135^{\circ}$

Where $\theta$ and $2 \theta-45^{\circ}$ are acute angles

Since $\theta=45^{\circ}$

Now

$\tan \theta$

$=\tan 45^{\circ}$ Put $\theta=45^{\circ}$

$=1$

Hence the correct option is $(a)$