Question : If $\sin\theta +\cos\theta=\sqrt{2}\sin(90^{\circ}-\theta)$, then the value of $\cot\theta$ is:
Option 1: $-\sqrt{2}-1$
Option 2: $\sqrt{2}-1$
Option 3: $\sqrt{2}+1$
Option 4: $-\sqrt{2}+1$
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Correct Answer: $\sqrt{2}+1$
Solution :
Given: $\sin\theta +\cos\theta=\sqrt{2}\sin(90^{\circ}-\theta)$
$⇒\sin\theta +\cos\theta=\sqrt{2}\cos\theta$
$⇒\sin\theta=(\sqrt{2}-1)\cos\theta$
$⇒\cot\theta=\frac{1}{\sqrt{2}-1}$
By rationalisation, $\cot\theta=\sqrt{2}+1$
Hence, the correct answer is $\sqrt{2}+1$.
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