Question
If $(\sin\theta+\cos\theta)=\sqrt{2}\cos\theta,$ show that $\cot\theta=\big(\sqrt{2}+1\big).$
✓
Answer
We have,
$(\sin\theta+\cos\theta)=\sqrt{2}\cos\theta$
Dividing both side by $\sin\theta,$ we get:
$\frac{\sin\theta}{\sin\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sqrt{2}\cos\theta}{\sin\theta}$
$\Rightarrow1+\cot\theta=\sqrt{2}\cot\theta$
$\Rightarrow\sqrt{2}\cot\theta-\cot\theta=1$
$\Rightarrow\Big(\sqrt{2}-1\Big)\cot\theta=1$
$\Rightarrow\cot\theta=\frac{1}{\big(\sqrt{2}-1\big)}$
$\Rightarrow\cot\theta=\frac{1}{\big(\sqrt{2}-1\big)}\times\frac{\big(\sqrt{2}+1\big)}{\big(\sqrt{2}+1\big)}$
$\Rightarrow\cot\theta=\frac{\big(\sqrt{2}+1\big)}{2-1}$
$\Rightarrow\cot\theta=\frac{\big(\sqrt{2}+1\big)}{1}$
$\therefore\ \cot\theta=\big(\sqrt{2}+1\big)$
$(\sin\theta+\cos\theta)=\sqrt{2}\cos\theta$
Dividing both side by $\sin\theta,$ we get:
$\frac{\sin\theta}{\sin\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sqrt{2}\cos\theta}{\sin\theta}$
$\Rightarrow1+\cot\theta=\sqrt{2}\cot\theta$
$\Rightarrow\sqrt{2}\cot\theta-\cot\theta=1$
$\Rightarrow\Big(\sqrt{2}-1\Big)\cot\theta=1$
$\Rightarrow\cot\theta=\frac{1}{\big(\sqrt{2}-1\big)}$
$\Rightarrow\cot\theta=\frac{1}{\big(\sqrt{2}-1\big)}\times\frac{\big(\sqrt{2}+1\big)}{\big(\sqrt{2}+1\big)}$
$\Rightarrow\cot\theta=\frac{\big(\sqrt{2}+1\big)}{2-1}$
$\Rightarrow\cot\theta=\frac{\big(\sqrt{2}+1\big)}{1}$
$\therefore\ \cot\theta=\big(\sqrt{2}+1\big)$
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