Question

If $sin\theta +cos\theta =\sqrt{2}sin({90}^{\circ }-\theta )$ determine cot $\theta .$

Answer 1

We have,
$sin\theta +cos\theta =\sqrt{2}sin({90}^{\circ }-\theta )$

$\Rightarrow sin\theta +cos\theta =\sqrt{2}cos\theta$

$\Rightarrow sin\theta =\sqrt{2}cos\theta -cos\theta$

$\Rightarrow sin\theta =(\sqrt{2}-1)cos\theta$

$\Rightarrow \frac{sin\theta }{cos\theta }=\frac{(\sqrt{2}-1)cos\theta }{cos\theta }$

$\Rightarrow tan\theta =(\sqrt{2}-1)$

$\Rightarrow cot\theta =\frac{1}{\sqrt{2}-1}[\because cot\theta =\frac{1}{tan\theta }]$

$\Rightarrow cot\theta =\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{(\sqrt{2}+1)}{2-1}=\sqrt{2}+1$