In-any-triangle-ABC-a-cot-A-b-cot-B-c-cot-C-is-equal-to-1-r-R-2-r-R-3-2-r-R-4-2-r-R-

Question Number 16358 by Tinkutara last updated on 21/Jun/17

In any triangle A B C, a \cot A + b \cot B

+ c \cot C is equal to

(1) r + R

(2) r - R

(3) 2 (r + R)

(4) 2 (r - R)

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Jun/17

a n s w e r (3) .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Jun/17

a c o t g A = \frac{2 R s i n A . c o s A}{s i n A} = 2 R c o s A

\Rightarrow L H S = 2 R (c o s A + c o s B + c o s C)

Σ c o s A = 2 c o s \frac{A + B}{2} . c o s \frac{A - B}{2} + c o s (180 - A - B) =

= 2 c o s \frac{A + B}{2} . c o s \frac{A - B}{2} - 2 {c o s}^{2} \frac{A + B}{2} + 1 =

2 c o s \frac{A + B}{2} (c o s \frac{A - B}{2} - c o s \frac{A + B}{2}) + 1 =

= 2 c o s \frac{180 - C}{2} . 2 s i n \frac{A}{2} s i n \frac{B}{2} + 1 =

= 4 s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2} + 1

s i n \frac{A}{2} = \sqrt{\frac{1 - c o s A}{2}} = \sqrt{\frac{1 - \frac{b^{2} + c^{2} - a^{2}}{2 b c}}{2}} =

= \sqrt{\frac{(a + b - c) (a + c - b)}{4 b c}} = \sqrt{\frac{(p - b) (p - c)}{b c}}

\Rightarrow Π s i n \frac{A}{2} = \sqrt{\frac{(p - b) (p - c) (p - a) (p - c) (p - a) (p - b)}{a^{2} b^{2} c^{2}}} =

= \frac{(p - a) (p - b) (p - c)}{a b c} = \frac{\frac{S^{2}}{p}}{4 R . S} = \frac{S}{4 R . p} = \frac{r}{4 R}

\Rightarrow L H S = 2 R (4 \times \frac{r}{4 R} + 1) = 2 r + 2 R = 2 (r + R) .

Commented by Tinkutara last updated on 22/Jun/17

Thanks Sir!

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

n o t e :

s i n A + s i n B + s i n C = 4 c o s \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2}

c o s \frac{A}{2} = \sqrt{\frac{1 + c o s A}{2}} = \sqrt{\frac{1 + \frac{b^{2} + c^{2} - a^{2}}{2 b c}}{2}} =

= \sqrt{\frac{(b + c + a) (b + c - a)}{4 b c}} = \sqrt{\frac{p (p - a)}{b c}}

\Rightarrow Π c o s \frac{A}{2} = \sqrt{\frac{p (p - a) p (p - b) p (p - c)}{a^{2} b^{2} c^{2}}} =

= \frac{p . S}{a b c} = \frac{p . S}{4 R . S} = \frac{p}{4 R}

t g A + t g B + t g (180 \overset{(= C)}{-} A - B) =

= t g A + t g B - \frac{t g A + t g B}{1 - t g A t g B} = (t g A + t g B) (1 - \frac{1}{1 - t g A . t g B}) =

= - t g A . t g B . \frac{t g A + t g B}{1 - t g A . t g B} = - t g A . t g B . (- t g C) =

= t g A . t g B . t g C

Σ s i n A = 4 Π c o s \frac{A}{2}

Σ c o s A = 1 + 4 Π s i n \frac{A}{2}

Σ t g A = Π t g A