​$$\text{Using } \cot(A - B) = \frac{\cot A \cdot \cot B + 1}{\cot B - \cot A}, \text{ find the value of } \cot 15^\circ.$$​​

Given: 

Trigonometric expression ; $$\cot(A - B) = \frac{\cot A \cdot \cot B + 1}{\cot B - \cot A}$$ 

To find: value of $$\cot 15^\circ$$ 

Formula Used:  

​$$\cot(A - B) = \frac{\cot A \cdot \cot B + 1}{\cot B - \cot A}$$​​

Solution: 

​$$\cot(A - B) = \frac{\cot A \cdot \cot B + 1}{\cot B - \cot A} \\\Rightarrow \cot 15^\circ = \cot (45^\circ - 30^\circ) \\\Rightarrow \cot (45^\circ - 30^\circ) = \frac{\cot 45^\circ \cdot \cot 30^\circ + 1}{\cot 30^\circ - \cot 45^\circ} \\\Rightarrow \frac{1 \times \sqrt{3} + 1}{\sqrt{3} - 1} \\\Rightarrow \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \\ \text{ multiplying denominator and numerator by} \sqrt3 +1 \\ \Rightarrow \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\\Rightarrow \frac{(\sqrt{3} + 1)^2}{2} \\\Rightarrow \frac{3 + 1 + 2\sqrt{3}}{2} \\\Rightarrow \frac{(2\sqrt{3} + 4)}{2} \\\Rightarrow \sqrt{3} + 2$$​​​​