Question

If sin θ + cos θ = $\sqrt{2}$ sin (90° − θ), show that cot θ = $\left(\sqrt{2}+1\right)$.

Answer 1

$$\begin{gathered} \text{Given: cos}\theta \text{+sin}\theta \text{=}\sqrt{2}\sin ({90}^0-\theta ) \\ \text{=>cos}\theta \text{+sin}\theta =\sqrt{2}\cos \theta \text{{∵ sin(90°-θ) = cosθ}} \\ \text{=>}\frac{\text{cos}\theta \text{+sin}\theta }{\sin \theta }=\frac{\sqrt{2}\cos \theta }{\sin \theta }\left(\text{Dividing both sides by sin}\theta \right) \\ \text{=>cot}\theta +1\text{=}\sqrt{2}\cot \theta \\ \text{=>}(\sqrt{2}-1)\cot \theta =1 \\ \text{=>cot}\theta \text{=}\frac{1}{(\sqrt{2}-1)} \\ \text{=>cot}\theta \text{=}\frac{1}{(\sqrt{2}-1)}\times \frac{(\sqrt{2}+1)}{(\sqrt{2}+1)} \\ \begin{array}{l}\text{=>cot}\theta \text{=}\frac{(\sqrt{2}+1)}{2-1}\\ \end{array} \\ \text{=>cot}\theta \text{=}(\sqrt{2}+1) \\ \text{Hence proved}\text{.} \end{gathered}$$