The value of cot(7.5^{circ}) + tan(67.5^{circ | vedclass

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(B) दी गई अभिव्यक्ति: $E = (\cot 7.5^{\circ} - 	an 7.5^{\circ}) + (	an 67.5^{\circ} - \cot 67.5^{\circ})$हम जानते हैं कि $\cot 	heta - 	an 	heta

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an irrational number

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(B) दी गई अभिव्यक्ति: $E = (\cot 7.5^{\circ} - 	an 7.5^{\circ}) + (	an 67.5^{\circ} - \cot 67.5^{\circ})$हम जानते हैं कि $\cot 	heta - 	an 	heta = \frac{\cos 	heta}{\sin 	heta} - \frac{\sin 	heta}{\cos 	heta} = \frac{\cos^2 	heta - \sin^2 	heta}{\sin 	heta \cos 	heta} = \frac{\cos 2	heta}{\frac{1}{2} \sin 2	heta} = 2 \cot 2	heta$अतः,पहला पद: $\cot 7.5^{\circ} - 	an 7.5^{\circ} = 2 \cot(2 	imes 7.5^{\circ}) = 2 \cot 15^{\circ}$दूसरा पद: $	an 67.5^{\circ} - \cot 67.5^{\circ} = -(\cot 67.5^{\circ} - 	an 67.5^{\circ}) = -2 \cot(2 	imes 67.5^{\circ}) = -2 \cot 135^{\circ}$यहाँ,$\cot 15^{\circ} = \cot(45^{\circ} - 30^{\circ}) = \frac{\cot 45^{\circ} \cot 30^{\circ} + 1}{\cot 30^{\circ} - \cot 45^{\circ}} = \frac{1 	imes \sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 1 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$और $\cot 135^{\circ} = \cot(180^{\circ} - 45^{\circ}) = -\cot 45^{\circ} = -1$अतः,$E = 2(2 + \sqrt{3}) - 2(-1) = 4 + 2\sqrt{3} + 2 = 6 + 2\sqrt{3} = 2(3 + \sqrt{3})$नोट: विकल्पों में सुधार के बाद,सही उत्तर $2(3 + \sqrt{3})$ है,जो एक अपरिमेय संख्या है।

The value of $\cot(7.5^{\circ}) + \tan(67.5^{\circ}) - \cot(67.5^{\circ}) - \tan(7.5^{\circ})$ is:

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