Show that \( \frac{\cos \left(45^{\circ}+\theta\right)}{\cos \left(45^{\circ}-\theta\right)}=\frac{1-\tan \theta}{1+\tan \theta} \)

To demonstrate that \[ \frac{\cos \left(45^{\circ}+\theta\right)}{\cos \left(45^{\circ}-\theta\right)} = \frac{1 - \tan \theta}{1 + \tan \theta}, \] we can use trigonometric identities and algebraic manipulation. Here's a step-by-step proof: ### 1. Expand the Cosine Terms Using the Angle Addition and Subtraction Formulas The cosine of a sum and difference can be expanded as: \[ \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B. \] Applying this to both the numerator and the denominator: - **Numerator:** \[ \cos(45^{\circ} + \theta) = \cos 45^{\circ} \cos \theta - \sin 45^{\circ} \sin \theta. \] - **Denominator:** \[ \cos(45^{\circ} - \theta) = \cos 45^{\circ} \cos \theta + \sin 45^{\circ} \sin \theta. \] ### 2. Substitute the Values for \(\cos 45^{\circ}\) and \(\sin 45^{\circ}\) We know that: \[ \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2}. \] Substitute these into the expressions: - **Numerator:** \[ \cos(45^{\circ} + \theta) = \frac{\sqrt{2}}{2} \cos \theta - \frac{\sqrt{2}}{2} \sin \theta = \frac{\sqrt{2}}{2} (\cos \theta - \sin \theta). \] - **Denominator:** \[ \cos(45^{\circ} - \theta) = \frac{\sqrt{2}}{2} \cos \theta + \frac{\sqrt{2}}{2} \sin \theta = \frac{\sqrt{2}}{2} (\cos \theta + \sin \theta). \] ### 3. Form the Ratio of Numerator to Denominator \[ \frac{\cos(45^{\circ} + \theta)}{\cos(45^{\circ} - \theta)} = \frac{\frac{\sqrt{2}}{2} (\cos \theta - \sin \theta)}{\frac{\sqrt{2}}{2} (\cos \theta + \sin \theta)} = \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}. \] ### 4. Simplify the Expression by Dividing Numerator and Denominator by \(\cos \theta\) Assuming \(\cos \theta \neq 0\), divide both the numerator and the denominator by \(\cos \theta\): \[ \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} = \frac{1 - \tan \theta}{1 + \tan \theta}, \] where \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). ### 5. Conclusion Thus, we have shown that: \[ \frac{\cos \left(45^{\circ}+\theta\right)}{\cos \left(45^{\circ}-\theta\right)} = \frac{1 - \tan \theta}{1 + \tan \theta}. \] **Final Answer:** After expanding, the ratio simplifies to (1 – tan θ) divided by (1 + tan θ). Thus, \[\frac{\cos(45°+θ)}{\cos(45°-θ)} = \frac{1 - \tan θ}{1 + \tan θ}.\]