Given-A-B-C-180-prove-that-tan-A-2-tan-B-2-tan-B-2-tan-C-2-tan-C-2-tan-A-2-1-

Question Number 99742 by Ar Brandon last updated on 23/Jun/20

G iven A + B + C = 180 ° prove that

\tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} = 1

Answered by smridha last updated on 23/Jun/20

\boldsymbol t a n (\frac{\boldsymbol A}{2} + \frac{\boldsymbol B}{2} + \frac{\boldsymbol C}{2}) = \boldsymbol t a n \frac{\boldsymbol π}{2} = \infty

\Rightarrow \frac{\boldsymbol t a n \frac{\boldsymbol A}{2} + \boldsymbol t a n \frac{\boldsymbol B}{2} + \boldsymbol t a n \frac{\boldsymbol C}{2} - \boldsymbol t a n \frac{\boldsymbol A}{2} . \boldsymbol t a n \frac{\boldsymbol B}{2} \boldsymbol t a n \frac{\boldsymbol C}{2}}{1 - [\boldsymbol t a n \frac{\boldsymbol A}{2} . \boldsymbol t a n \frac{\boldsymbol B}{2} + \boldsymbol t a n \frac{\boldsymbol B}{2} . \boldsymbol t a n \frac{\boldsymbol C}{2} + \boldsymbol t a n \frac{\boldsymbol C}{2} . \boldsymbol t a n \frac{\boldsymbol A}{2}]} = \infty

\boldsymbol t h i s \boldsymbol i s \boldsymbol p o s s i b l e \boldsymbol w h e n \boldsymbol t h e \boldsymbol d e n o m i n a t o r

\boldsymbol g o e s \boldsymbol t o 0

\boldsymbol t h e r e f o r e . .

[\boldsymbol t a n \frac{\boldsymbol A}{2} \boldsymbol t a n \frac{\boldsymbol B}{2} + \boldsymbol t a n \frac{\boldsymbol B}{2} . \boldsymbol t a n \frac{\boldsymbol C}{2} + \boldsymbol t a n \frac{\boldsymbol C}{2} . \boldsymbol t a n \frac{\boldsymbol A}{2}] = 1

Commented by smridha last updated on 23/Jun/20

\boldsymbol i s \boldsymbol t h e r e \boldsymbol a n y \boldsymbol l o g i c \boldsymbol b e h i n d \boldsymbol y o u r

\boldsymbol s i l l y \boldsymbol q u e s t i o n ? ?

Commented by Ar Brandon last updated on 23/Jun/20

Thank you Sir

Commented by smridha last updated on 23/Jun/20

yeah!! welcome

Commented by Rasheed.Sindhi last updated on 23/Jun/20

S i r, i s t h e r e a n y l o g i c b e h i n d

w r i t i n g i n d i f f e r e n t c o l o u r s ?

Commented by Rasheed.Sindhi last updated on 23/Jun/20

I f y o u d o s o m e t h i n g d i f f e r e n t,

y o u s h o u l d h a v e a r e a s o n f o r t h a t .

S p e c i a l l y i f y o u a r e m a t h - r e l a t e d

p e r s o n . M a y b e \boldsymbol o n l y \boldsymbol b e a t i f i c a t i o n

b e a r e a s o n .

Commented by smridha last updated on 23/Jun/20

\boldsymbol w h y \boldsymbol o b s e s s e d \boldsymbol f o r \boldsymbol b l a c k \boldsymbol a n d \boldsymbol w h i t e

\boldsymbol m a n!! \boldsymbol t h e r e \boldsymbol i s \boldsymbol a \boldsymbol f a c i l i t y \boldsymbol f o r

\boldsymbol c h o o s i n g \boldsymbol d e f r e n t \boldsymbol c o l o u r s . . \boldsymbol y o u \boldsymbol m a y

\boldsymbol u s e \boldsymbol o r \boldsymbol y o u \boldsymbol m a y \boldsymbol n o t . . \boldsymbol {t h a t}^{'} s \boldsymbol y o u r

\boldsymbol c h o i c e \dots \boldsymbol w h i c h \boldsymbol c o l o u r \boldsymbol c o m b i n a t i o n

I \boldsymbol u s e d \boldsymbol {t h a t}^{'} s \boldsymbol m y \boldsymbol b u s i n e s s \dots

\boldsymbol t h e \boldsymbol s o l u t i o n \boldsymbol i s \boldsymbol p r o m i n e n t \boldsymbol a n d

\boldsymbol b o l t \boldsymbol e n o u g h \boldsymbol s o \boldsymbol t h e r e \boldsymbol s h a l l \boldsymbol b e \boldsymbol n o

\boldsymbol o b j e c t i o n \dots .

Commented by Rasheed.Sindhi last updated on 23/Jun/20

O k \boldsymbol s I R a S Y ⊚_{\boldsymbol U} W i S^{h}!

I o n l y w a n t t o s a y t h a t u s e o f

d i f f e r e n t c o l o r s s h o u l d b e h e l p f u l

i n u n d e r s t a n d i n g p r o c e s s . I t s h o u l d

b e u s e d a s a t o o l .

Commented by Ar Brandon last updated on 23/Jun/20

��cheers

Commented by Rasheed.Sindhi last updated on 24/Jun/20

��cool !

Commented by Rasheed.Sindhi last updated on 24/Jun/20

S i r / M a d a m S m r i d h a

S o r r y i f m y c o o o l w o r d s m a d e y o u

\boldsymbol a n g r y . A c t u a l l y I {d i d n}^{'} t w a n t t h a t .

Commented by smridha last updated on 24/Jun/20

\boldsymbol t h i s \boldsymbol i s \boldsymbol n o t \boldsymbol f o r \boldsymbol c o o l \boldsymbol w o r d s

\boldsymbol t h i s \boldsymbol i s \boldsymbol f o r \boldsymbol y o u r \boldsymbol p r e v i o u s \boldsymbol c o m m e n t .

\boldsymbol t h i s \boldsymbol w i l l \boldsymbol b e \boldsymbol g o o d \boldsymbol i f \boldsymbol y o u \boldsymbol f o c u s

\boldsymbol o n \boldsymbol s o l v i n g \boldsymbol p r o b l e m \boldsymbol i n s t e a d \boldsymbol o f

\boldsymbol c r e a t i n g \boldsymbol v i o l e n c e . \boldsymbol I d o \boldsymbol n o t \boldsymbol k n o w

\boldsymbol h o w \boldsymbol l o n g \boldsymbol y o u \boldsymbol a r e \boldsymbol p r e s e n t \boldsymbol i n \boldsymbol t h i s

\boldsymbol f o r u m \dots \boldsymbol b u t \boldsymbol h o w \boldsymbol m u c h \boldsymbol I \boldsymbol k n o w

\boldsymbol t h e r e \boldsymbol i s \boldsymbol n o \boldsymbol m a d a m!! \boldsymbol {t h a t}^{'} \boldsymbol s \boldsymbol y o u r

VERDENCY \dots

Commented by Rasheed.Sindhi last updated on 24/Jun/20

\boldsymbol S i r S m r i d h a, p l g o a n d w o r k

s m o o t h l y . W r i t e y o u r c o l o r f u l /

b e a u t i f u l a n s w e r s . N o o n e w i l l

d i s t u r b y o u! . . W h e n I s c r o l l t h r o u g h

p o s t s o f t h e f o r u m, I i m e d i a t e l y

r e c o g n i s e y o r p o s t s w i t h o u t r e a d i n g

y o u r n a m e! \dots .

A c t u a l l y I {w a s n}^{'} t c e r t a i n o f y o u r

b e i n g M r . s o I w r i t e b o t h s i r &

m a d a m . T h e r e w a s n o o t h e r

r e a s o n . {D o n}^{'} t t h i n k n e g a t i v e l y

f o r o t h e r s . T h a n k y o u . B y e!

(T h e t h r e a d h a s b e e n t o o l o n g!)

Answered by Dwaipayan Shikari last updated on 23/Jun/20

t a n \frac{B}{2} [\frac{s i n \frac{(A + C)}{2}}{c o s \frac{C}{2} c o s \frac{A}{2}}] + \frac{s i n \frac{A}{2} s i n \frac{C}{2}}{c o s \frac{A}{2} c o s \frac{C}{2}} {{{\frac{s i n \frac{A}{2}}{c o s \frac{A}{2}} + \frac{s i n \frac{C}{2}}{c o s \frac{C}{2}} = \frac{s i n \frac{(A + C)}{2}}{c o s \frac{A}{2} c o s \frac{C}{2}}}}}

= \frac{s i n (\frac{π}{2} - \frac{A + C}{2}) + s i n \frac{A}{2} s i n \frac{C}{2}}{c o s \frac{A}{2} c o s \frac{C}{2}} = \frac{c o s \frac{A}{2} c o s \frac{C}{2}}{c o s \frac{A}{2} c o s \frac{C}{2}} = 1 [P r o v e d] {{{B e c a u s e t a n \frac{A}{2} t a n \frac{B}{2} + t a n \frac{C}{2} t a n \frac{B}{2} = t a n \frac{B}{2} (\frac{s i n (A + C)}{c o s \frac{A}{2} c o s \frac{C}{2}})

{s i n \frac{A + C}{2} = c o s \frac{B}{2}

{s i n (\frac{π}{2} - \frac{A + C}{2})

= c o s \frac{A}{2} c o s \frac{C}{2} - s i n \frac{A}{2} s i n \frac{C}{2}

Commented by Ar Brandon last updated on 23/Jun/20

Thank you Sir

Commented by Dwaipayan Shikari last updated on 23/Jun/20

please don't tell me sir .I am a student. I am in pleasure to solve your problem . Thanking you.

Commented by Ar Brandon last updated on 23/Jun/20

As you wish ! ��

Answered by 1549442205 last updated on 23/Jun/20

We have \frac{A + B}{2} = \frac{π}{2} - \frac{C}{2} \Rightarrow \tan \frac{A + B}{2} = \tan (\frac{π}{2} - \frac{C}{2})

\Rightarrow \frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{1 - \tan \frac{A}{2} \tan \frac{B}{2}} = \cot \frac{C}{2} = \frac{1}{\tan \frac{C}{2}}

\Rightarrow \tan \frac{A}{2} \tan \frac{C}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{A}{2} \tan \frac{B}{2} = 1 (q . e . d)

Commented by Ar Brandon last updated on 23/Jun/20

wow ! that was brilliant. Thanks

Commented by 1549442205 last updated on 25/Jun/20

you are wellcome, sir!

you are wellcome, sir!