Find-the-type-of-triangle-such-that-the-following-relationship-holds-between-its-angles-tan-B-tan-C-tan-2-B-C-2-

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