If (1 + tanδ) = $\sqrt{2}$, then which of the following options corresponds to (cotδ − 1)?

1 + tanδ = $\sqrt{2}$ 

⇒ $1+\frac{\sin \delta }{\cos \delta }=\sqrt{2}$

⇒ cosδ + sinδ = $\sqrt{2}$ cosδ … (i)

Now, (cosδ+ sinδ)² + (cosδ − sinδ)² = 2

($\sqrt{2}$cosδ)² + (cosδ − sinδ)² = 2 

⇒ (cosδ − sinδ)² = 2 − 2cos²δ

⇒ (cosδ − sinδ) = $\sqrt{2\left(1-{\cos }^2\delta \right)}$ 

⇒ cosδ − sinδ = $\sqrt{2{\sin }^2\delta }$

⇒ $\frac{\cos \delta }{\sin \delta }-\frac{\sin \delta }{\sin \delta }=\sqrt{2}$

⇒ cotδ − 1 = $\sqrt{2}$