यदि tan {3^o} + 2tan {6^o} + 4tan {12^o} + 8c | vedclass

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(B) हम सर्वसमिका $\cot \alpha - 	an \alpha = 2 \cot 2\alpha$ का उपयोग करते हैं,जिसका अर्थ है $\cot \alpha = 	an \alpha + 2 \cot 2\alpha$.दी गई अभिव्

Vedclass · Accepted Answer

$\cot (15	heta)^o = 1$

Vedclass · Answer

(B) हम सर्वसमिका $\cot \alpha - 	an \alpha = 2 \cot 2\alpha$ का उपयोग करते हैं,जिसका अर्थ है $\cot \alpha = 	an \alpha + 2 \cot 2\alpha$.दी गई अभिव्यक्ति: $E = 	an 3^{\circ} + 2 	an 6^{\circ} + 4 	an 12^{\circ} + 8 \cot 24^{\circ}$.$	an \alpha = \cot \alpha - 2 \cot 2\alpha$ का उपयोग करने पर:$E = (\cot 3^{\circ} - 2 \cot 6^{\circ}) + 2 	an 6^{\circ} + 4 	an 12^{\circ} + 8 \cot 24^{\circ}$.चूंकि $2 	an 6^{\circ} - 2 \cot 6^{\circ} = -2(2 \cot 12^{\circ}) = -4 \cot 12^{\circ}$:$E = \cot 3^{\circ} - 4 \cot 12^{\circ} + 4 	an 12^{\circ} + 8 \cot 24^{\circ}$.चूंकि $4 	an 12^{\circ} - 4 \cot 12^{\circ} = -4(2 \cot 24^{\circ}) = -8 \cot 24^{\circ}$:$E = \cot 3^{\circ} - 8 \cot 24^{\circ} + 8 \cot 24^{\circ} = \cot 3^{\circ}$.अतः,$\cot 	heta^{\circ} = \cot 3^{\circ}$,इसलिए $	heta = 3$.विकल्पों की जाँच करने पर: $\cot (15 	imes 3)^{\circ} = \cot 45^{\circ} = 1$.

यदि $\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ है,तो:

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