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Question Number 119548 by benjo_mathlover last updated on 25/Oct/20

Commented by benjo_mathlover last updated on 25/Oct/20

$I f \cos A = \tan B, \cos B = \tan C a n d$ $\cos C = \tan A, t h e n \sin A i s e q u a l t o$
Commented by TANMAY PANACEA last updated on 25/Oct/20

${c o s}^{2} A = {t a n}^{2} B = {s e c}^{2} B - 1$ ${c o s}^{2} A = \frac{1}{{c o s}^{2} B} - 1$ ${c o s}^{2} A = \frac{1}{{t a n}^{2} C} - 1 [g i v e n c o s B = t a n C]$ ${c o s}^{2} A = \frac{1}{{s e c}^{2} C - 1} - 1 = \frac{1}{\frac{1}{{c o s}^{2} C} - 1} - 1 = \frac{{c o s}^{2} C}{1 - {c o s}^{2} C} - 1$ ${c o s}^{2} A = \frac{{c o s}^{2} C - 1 + {c o s}^{2} C}{1 - {c o s}^{2} C} = \frac{2 {t a n}^{2} A - 1}{1 - {t a n}^{2} A}$ ${c o s}^{2} A = \frac{2 {s i n}^{2} A - {c o s}^{2} A}{{c o s}^{2} A - {s i n}^{2} A}$ $s i n A = k$ $1 - k^{2} = \frac{2 k^{2} - (1 - k^{2})}{(1 - k^{2}) - k^{2}} = \frac{3 k^{2} - 1}{1 - 2 k^{2}}$ $k^{2} = 1 - \frac{3 k^{2} - 1}{1 - 2 k^{2}} = \frac{1 - 2 k^{2} - 3 k^{2} + 1}{1 - 2 k^{2}}$ $k^{2} - 2 k^{4} = 2 - 5 k^{2}$ $2 k^{4} - 6 k^{2} + 2 = 0$ $k^{4} - 3 k^{2} + 1 = 0$ $k^{2} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$ $k^{2} = \frac{6 \pm 2 \sqrt{5}}{4} = {(\frac{\sqrt{5} \pm 1}{2})}^{2}$ $k = s i n A = \frac{\sqrt{5} - 1}{2}$
	Answered by bemath last updated on 25/Oct/20
	$let sin A = k = tan B → { ((tan A=(k/( (√(1−k^2 )))))),((cos B = (1/( (√(1+k^2 )))) = tan C)) :} → cos A cos B = sin B → cos B cos C = sin C → cos A cos C = sin A (1)×(2)×(3) →(cos A cos B cos C )^2 = sin A sin B sin C ((√(1−k^2 )) . (1/( (√(1+k^2 )))). ((√(1+k^2 ))/( (√(2+k^2 )))) )^2 = k.(k/( (√(1+k^2 )))). (1/( (√(2+k^2 )))) → ((1−k^2 )/(2+k^2 )) = (k^2 /( (√((1+k^2 )(2+k^2 ))))) let k^2 = u (√((2+u)(1+u))) = (((2+u)u)/(1−u)) (2+u)(1+u)= ((u^2 (2+u)^2 )/((1−u)^2 )) (2+u){(1+u)(1−u)^2 −u^2 (2+u)} = 0 (2+u){ (1−u^2 )(1−u)−2u^2 −u^3 } = 0 (2+u) { 1−u−u^2 +u^3 −2u^2 −u^3 } = 0 (2+u) { 1−u−3u^2 } = 0 u=−2 ←rejected → 3u^2 +u−1 = 0 ⇒ u = ((−1+(√(13)))/6) = k^2 k = (√(((√(13))−1)/6))$
	$l e t \sin A = k = \tan B$ $\to {\begin{cases} \tan A = \frac{k}{\sqrt{1 - k^{2}}} \\ \cos B = \frac{1}{\sqrt{1 + k^{2}}} = \tan C \end{cases}$ $\to \cos A \cos B = \sin B$ $\to \cos B \cos C = \sin C$ $\to \cos A \cos C = \sin A$ $(1) \times (2) \times (3)$ $\to {(\cos A \cos B \cos C)}^{2} = \sin A \sin B \sin C$ ${(\sqrt{1 - k^{2}} . \frac{1}{\sqrt{1 + k^{2}}} . \frac{\sqrt{1 + k^{2}}}{\sqrt{2 + k^{2}}})}^{2} = k . \frac{k}{\sqrt{1 + k^{2}}} . \frac{1}{\sqrt{2 + k^{2}}}$ $\to \frac{1 - k^{2}}{2 + k^{2}} = \frac{k^{2}}{\sqrt{(1 + k^{2}) (2 + k^{2}})}$ $l e t k^{2} = u$ $\sqrt{(2 + u) (1 + u)} = \frac{(2 + u) u}{1 - u}$ $(2 + u) (1 + u) = \frac{u^{2} {(2 + u)}^{2}}{{(1 - u)}^{2}}$ $(2 + u) {(1 + u) {(1 - u)}^{2} - u^{2} (2 + u)} = 0$ $(2 + u) {(1 - u^{2}) (1 - u) - 2 u^{2} - u^{3}} = 0$ $(2 + u) {1 - u - u^{2} + u^{3} - 2 u^{2} - u^{3}} = 0$ $(2 + u) {1 - u - 3 u^{2}} = 0$ $u = - 2 \leftarrow r e j e c t e d$ $\to 3 u^{2} + u - 1 = 0$ $\Rightarrow u = \frac{- 1 + \sqrt{13}}{6} = k^{2}$ $k = \sqrt{\frac{\sqrt{13} - 1}{6}}$
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