1. Stating the problem: We want to verify the trigonometric identity $$\tan(90^\circ + \theta) = -\cot \theta$$. 2. Recall the definition and properties: - The tangent of an angle is defined as $$\tan x = \frac{\sin x}{\cos x}$$. - The cotangent is defined as $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$. - Use the angle addition formula or co-function identities for $$\tan(90^\circ + \theta)$$. 3. Use the identity for tangent of sum: $$\tan(90^\circ + \theta) = \frac{\tan 90^\circ + \tan \theta}{1 - \tan 90^\circ \tan \theta}$$. But $$\tan 90^\circ$$ is undefined (approaches infinity), so instead use the co-function identity: $$\tan(90^\circ + \theta) = -\cot \theta$$ directly from the trigonometric co-function property. 4. To verify mathematically: Since $$\tan(90^\circ + \theta) = \tan\left(\frac{\pi}{2} + \theta\right)$$ in radians, consider the sine and cosine: $$\tan\left(\frac{\pi}{2} + \theta\right) = \frac{\sin(\frac{\pi}{2} + \theta)}{\cos(\frac{\pi}{2} + \theta)}$$ Using angle sum: $$\sin\left(\frac{\pi}{2} + \theta\right) = \sin \frac{\pi}{2} \cos \theta + \cos \frac{\pi}{2} \sin \theta = 1 \cdot \cos \theta + 0 = \cos \theta$$ $$\cos\left(\frac{\pi}{2} + \theta\right) = \cos \frac{\pi}{2} \cos \theta - \sin \frac{\pi}{2} \sin \theta = 0 - 1 \cdot \sin \theta = -\sin \theta$$ Therefore, $$\tan\left(\frac{\pi}{2} + \theta\right) = \frac{\cos \theta}{-\sin \theta} = -\frac{\cos \theta}{\sin \theta} = -\cot \theta$$ 5. Conclusion: The identity $$\tan(90^\circ + \theta) = -\cot \theta$$ is true and proven. Final answer: $$\tan(90^\circ + \theta) = -\cot \theta$$