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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Question : If $0\leq\theta\leq 90^{\circ}$ and $4\cos^{2}\theta-4\sqrt{3}\cos\theta+3=0$, then the value of $\theta$ is:
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