If $\theta$ and $2 \theta-45^{\circ}$ are acute angles such that $\sin \theta=\cos \left(2 \theta-45^{\circ}\right)$, then $\tan \theta$ is equal to
(a) 1
(b) $-1$
(C) $\sqrt{3}$
(d) $\frac{1}{\sqrt{3}}$
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 $(a)$
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