Question
If $\sin\theta+\cos\theta=\sqrt{2}\cos(90^\circ-\theta),$ find $\cot\theta$.
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Answer
$\sin\theta+\cos\theta=\sqrt{2}\cos(90^\circ-\theta)$
$\Rightarrow\ \sin\theta+\cos\theta=\sqrt{2}\sin\theta$
$\Rightarrow\ \cos\theta=\sqrt{2}\sin\theta-\sin\theta$
$\Rightarrow\ \cos\theta=\sin\theta(\sqrt{2}-1)$
$\Rightarrow\ \frac{\cos\theta}{\sin\theta}=(\sqrt{2}-1)$
$\Rightarrow\ \cot\theta=(\sqrt{2}-1)$
$\Rightarrow\ \sin\theta+\cos\theta=\sqrt{2}\sin\theta$
$\Rightarrow\ \cos\theta=\sqrt{2}\sin\theta-\sin\theta$
$\Rightarrow\ \cos\theta=\sin\theta(\sqrt{2}-1)$
$\Rightarrow\ \frac{\cos\theta}{\sin\theta}=(\sqrt{2}-1)$
$\Rightarrow\ \cot\theta=(\sqrt{2}-1)$
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