الحل: لحساب \(\tan 75^\circ\)، نستخدم صيغة جمع الزوايا للدالة الظل:! 🏅 \tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} - Get complete information, detailed analysis, and expert insights. Explore detailed breakdown with verified data and expert analysis. Get comprehensive coverage of all important aspects with latest information. Understand key facts, important details, and complete overview with professional insights. Learn about background, current status, future prospects, and everything you need to know. Discover in-depth analysis, expert commentary, and detailed explanations covering all relevant topics. WARNING: Explicit Content!

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الحل: لحساب \(\tan 75^\circ\)، نستخدم صيغة جمع الزوايا للدالة الظل:

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\[

\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}

\]

نعبر عن \(75^\circ\) كمجموع \(45^\circ + 30^\circ\):

\[

\tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}

\]

نعوض بالقيم المعروفة \(\tan 45^\circ = 1\) و \(\tan 30^\circ = \frac{1}{\sqrt{3}}\):

\[

\tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

\]

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