tan frac{2 pi}{7} cdot tan frac{4 pi}{7} + ta | vedclass

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Question

(B) Let $	heta = \frac{\pi}{7}$. Then $7	heta = \pi$,so $4	heta = \pi - 3	heta$. Taking tangent on both sides,$	an(4	heta) = 	an(\pi - 3	heta)

Vedclass · Accepted Answer

$-7$

Vedclass · Answer

(B) Let $	heta = \frac{\pi}{7}$. Then $7	heta = \pi$,so $4	heta = \pi - 3	heta$. Taking tangent on both sides,$	an(4	heta) = 	an(\pi - 3	heta) = -	an(3	heta)$. Using the expansion formulas: $\frac{4	an	heta - 4	an^3	heta}{1 - 6	an^2	heta + 	an^4	heta} = -\frac{3	an	heta - 	an^3	heta}{1 - 3	an^2	heta}$. Dividing by $	an	heta$ (since $	an	heta 
eq 0$): $4(1 - 	an^2	heta)(1 - 3	an^2	heta) = -(1 - 6	an^2	heta + 	an^4	heta)(3 - 	an^2	heta)$. Let $x = 	an^2	heta$. The equation becomes $4(1 - 4x + 3x^2) = -(3 - 19x + 9x^2 - x^3)$. $4 - 16x + 12x^2 = -3 + 19x - 9x^2 + x^3$. $x^3 - 21x^2 + 35x - 7 = 0$. The roots of this equation are $	an^2(\frac{\pi}{7}), 	an^2(\frac{2\pi}{7}), 	an^2(\frac{3\pi}{7})$. Using the identity $\sum 	an \alpha 	an \beta = -7$ for the roots of the related equation $	an(7	heta)=0$,the value of the expression is $-7$.

List-$I$	List-$II$
$(I)$ $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \dots + \sin^2 90^{\circ}$	$(A)$ $0$
$(II)$ $\tan^2 5^{\circ} \cdot \tan^2 10^{\circ} \cdot \tan^2 15^{\circ} \dots \tan^2 85^{\circ}$	$(B)$ $\frac{19}{2}$
$(III)$ $\cos^2 5^{\circ} + \cos^2 10^{\circ} + \cos^2 15^{\circ} + \dots + \cos^2 180^{\circ}$	$(C)$ $18$
$(IV)$ $\cot 5^{\circ} + \cot 10^{\circ} + \cot 15^{\circ} + \dots + \cot 175^{\circ}$	$(D)$ $1$
	$(E)$ $-1$

$\tan \frac{2 \pi}{7} \cdot \tan \frac{4 \pi}{7} + \tan \frac{4 \pi}{7} \cdot \tan \frac{\pi}{7} + \tan \frac{\pi}{7} \cdot \tan \frac{2 \pi}{7} = $

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