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

Question:

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) $\frac{1}{\sqrt{3}}$

Solution:

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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