The value of cot 7frac{1}{2}^{circ} + tan 67f | vedclass

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(B) Let the expression be $E = (\cot 7.5^{\circ} - 	an 7.5^{\circ}) - (\cot 67.5^{\circ} - 	an 67.5^{\circ})$.Using the identity $\cot 	heta - 	an

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an irrational number

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(B) Let the expression be $E = (\cot 7.5^{\circ} - 	an 7.5^{\circ}) - (\cot 67.5^{\circ} - 	an 67.5^{\circ})$.Using the identity $\cot 	heta - 	an 	heta = \frac{\cos 	heta}{\sin 	heta} - \frac{\sin 	heta}{\cos 	heta} = \frac{\cos^2 	heta - \sin^2 	heta}{\sin 	heta \cos 	heta} = \frac{\cos 2	heta}{\frac{1}{2} \sin 2	heta} = 2 \cot 2	heta$.For the first part,$	heta = 7.5^{\circ}$,so $2	heta = 15^{\circ}$. Thus,$\cot 7.5^{\circ} - 	an 7.5^{\circ} = 2 \cot 15^{\circ}$.For the second part,$	heta = 67.5^{\circ}$,so $2	heta = 135^{\circ}$. Thus,$\cot 67.5^{\circ} - 	an 67.5^{\circ} = 2 \cot 135^{\circ}$.Substituting these into $E$,we get $E = 2 \cot 15^{\circ} - 2 \cot 135^{\circ}$.We know $\cot 15^{\circ} = \cot(45^{\circ} - 30^{\circ}) = \frac{\cot 45^{\circ} \cot 30^{\circ} + 1}{\cot 30^{\circ} - \cot 45^{\circ}} = \frac{1 \cdot \sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.We know $\cot 135^{\circ} = -1$.Therefore,$E = 2(2 + \sqrt{3}) - 2(-1) = 4 + 2\sqrt{3} + 2 = 6 + 2\sqrt{3} = 2(3 + \sqrt{3})$.Wait,re-evaluating the expression: $E = 2 \cot 15^{\circ} - 2 \cot 135^{\circ} = 2(2 + \sqrt{3}) - 2(-1) = 4 + 2\sqrt{3} + 2 = 6 + 2\sqrt{3}$.Comparing with options,$2(3 + \sqrt{3})$ is not explicitly listed,but $2(3 + 2\sqrt{3})$ is option $C$. Let's re-check the question expression: $\cot 7.5^{\circ} - 	an 7.5^{\circ} = 2 \cot 15^{\circ}$. $	an 67.5^{\circ} - \cot 67.5^{\circ} = -2 \cot 135^{\circ} = 2$. So $E = 2(2+\sqrt{3}) + 2 = 6 + 2\sqrt{3} = 2(3+\sqrt{3})$. Since $2(3+\sqrt{3})$ is not an option,and $2(3+2\sqrt{3})$ is,there might be a typo in the question's options. However,$6+2\sqrt{3}$ is an irrational number. Thus,option $B$ is correct.

The value of $\cot 7\frac{1}{2}^{\circ} + \tan 67\frac{1}{2}^{\circ} - \cot 67\frac{1}{2}^{\circ} - \tan 7\frac{1}{2}^{\circ}$ is:

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