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Sure, let's go through the problems step by step. **1. Compute:** a) \( \tan 45^\circ - \sin 30^\circ \) - \( \tan 45^\circ = 1 \) - \( \sin 30^\circ = 0.5 \) So, \( 1 - 0.5 = 0.5 \). b) \( \cos 60^\circ + \tan 45^\circ \) - \( \cos 60^\circ = 0.5 \) - \( \tan 45^\circ = 1 \) So, \( 0.5 + 1 = 1.5 \). **2. Simplify the expressions:** a) \( (1 + \sin \alpha)(1 - \sin \alpha) \) This is a difference of squares: \[ 1 - \sin^2 \alpha = \cos^2 \alpha \] b) \( \tan \alpha \cdot \cos \alpha \) \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \] \[ \tan \alpha \cdot \cos \alpha = \sin \alpha \] c) \( 1 + \cos^2 \alpha - \sin^2 \alpha \) Using the identity \( \cos^2 \alpha = 1 - \sin^2 \alpha \): \[ 1 + (1 - \sin^2 \alpha) - \sin^2 \alpha = 2 - 2\sin^2 \alpha = 2\cos^2 \alpha \] **3. Find the value:** a) \( \cos \alpha \), if \( \sin \alpha = 0.6 \) Using \( \sin^2 \alpha + \cos^2 \alpha = 1 \): \[ 0.6^2 + \cos^2 \alpha = 1 \] \[ 0.36 + \cos^2 \alpha = 1 \] \[ \cos^2 \alpha = 0.64 \] \[ \cos \alpha = \sqrt{0.64} = 0.8 \] b) \( \sin \alpha \), if \( \cos \alpha = 0.8 \) Similarly: \[ \sin^2 \alpha + 0.8^2 = 1 \] \[ \sin^2 \alpha + 0.64 = 1 \] \[ \sin^2 \alpha = 0.36 \] \[ \sin \alpha = \sqrt{0.36} = 0.6 \] I hope this helps! Let me know if you have further questions.