\( \tan 75^\circ = ? \)

Another Explanation (5):

প্রশ্ন:

\( \tan 75^\circ = ? \)

উত্তর:

\( \frac{\sqrt{3}+1}{\sqrt{3}-1} \)

সমাধান:

আমরা জানি যে, \(\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)

এখানে, \(75^\circ = 45^\circ + 30^\circ\)

তাই,

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

পাশে, \(\tan 45^\circ = 1\) এবং \(\tan 30^\circ = \frac{1}{\sqrt{3}}\)

অতএব,

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

সাধারণীকরণে, উভয় উপরের ভগ্নাংশের জন্য কমন ডেনমিনেটর \(\sqrt{3}\):

\[ = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

অতএব,

উত্তর:

\[ \boxed{\frac{\sqrt{3} + 1}{\sqrt{3} - 1}} \]