Find the type of triangle such that the following relationship holds between its angles. tan (B)tan(C)= tan^2 (((B+C)/2) ) ■


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Question Number 215349 by mnjuly1970 last updated on 03/Jan/25

$F i n d t h e t y p e o f t r i a n g l e$ $s u c h t h a t t h e f o l l o w i n g$ $r e l a t i o n s h i p h o l d s b e t w e e n$ $i t s a n g l e s .$ $t a n (B) t a n (C) = {t a n}^{2} (\frac{B + C}{2}) ◼$
	Answered by MrGaster last updated on 04/Jan/25

	$\tan (B) \tan (C) = \tan^{2} (\frac{B + C}{2})$ $= \tan^{2} (\frac{π - A}{2})$ $= \cot^{2} (\frac{A}{2})$ $\tan (B) \tan (C) = {(\frac{\cos (\frac{A}{2})}{\sin (\frac{A}{2})})}^{2}$ $= (\frac{1 + \cos (A)}{1 - \cos (A)})$ $\tan (B) \tan (C) = {(\frac{\cos (\frac{A}{2})}{\sin {(\frac{A}{2})}^{2}})}^{2}$ $= (\frac{1 + \cos (A)}{1 - \cos (A)})$ $\tan (B) \tan (C) = (\frac{1 + \cos (B + C)}{1 - \cos (B + C)})$ $= (\frac{1 + (\cos (B) \cos (C) - \sin (B) \sin (C)}{1 - (\cos (B) \cos (C) - \sin (B) \sin (C))})$ $= (\frac{\cos (B) \cos (C) + \sin (B) \sin (C) + \sin (B) \sin (C))}{(\cos (B) \cos (A) + \sin (B) \sin (C) - \sin (B) \sin (C)})$ $= (\frac{\cos (B - C) + \sin (B) \sin (C)}{\cos (B - C)})$ $= 1 + \tan (B) \tan (C)$ $\tan (B) \tan (C) = 1 + \tan (B) \tan (C)$ $\tan (B) \tan (C) = \tan^{2} (\frac{B + C}{2})$ $= \tan^{2} (\frac{π - A}{2})$ $= \cot^{2} (\frac{A}{2})$ $\tan (B) \tan (C) = 1$ $\tan (B) \tan (C) = \tan (\frac{π}{4}) \tan (\frac{π}{4})$ $B = C = \frac{π}{4}$ $A = π - (B + C) = \frac{π}{2}$ $∴ △ A B C is a right isosecles triangle$
	Commented by mnjuly1970 last updated on 05/Jan/25

	$t h x s i r$
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