In a DeltaABC, prove that the value of cosA/(sinBsinC)+cosB/(sinCsinA)+cosC/(sinAsinB) is a prime?

Jul 17, 2018

$2$ and $2$ is a prime

Explanation:

Proof: we need the theorem of cosines
$\cos \left(\alpha\right) = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}$ etc

and the area of a triangle

$A = \frac{1}{2} a \cdot b \sin \left(\gamma\right)$ etc and the formula by Heron

$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ where $s = \frac{a + b + c}{2}$ so we get

$\frac{\frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}}{2 \frac{A}{a c} \cdot 2 \frac{A}{a b}} + \frac{\frac{{a}^{2} + {c}^{2} - {b}^{2}}{2 a c}}{2 \frac{A}{a b} \cdot 2 \frac{A}{b c}} + \frac{\frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}}{2 \frac{A}{b c} \cdot 2 \frac{A}{a c}}$

simplifying and expanding

$2 \cdot \frac{2 {a}^{2} {b}^{2} + 2 {a}^{2} {c}^{2} + 2 {b}^{2} {c}^{2} - {a}^{4} - {b}^{4} - {c}^{4}}{\left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right)}$
Now it is
$\left(a + b + c\right) \left(- a + b + c\right) \left(a - b + c\right) \left(a + b - c\right) = 2 {a}^{2} {b}^{2} + 2 {a}^{2} {c}^{2} + 2 {b}^{2} {c}^{2} - {a}^{4} - {b}^{4} - {c}^{4}$
so our result is $2$

Jul 17, 2018

Explanation:

We know that ,

$\left({F}_{1}\right) \sin \left(x + y\right) = 2 \sin \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)$

$\left({F}_{2}\right) \cos \left(x - y\right) = - 2 \sin \left(\frac{x + y}{2}\right) \sin \left(\frac{x - y}{2}\right)$

We take ,

X=cosA/(sinBsinC)+cosB/(sinCsinA)+cosC/(sinAsinB

$\implies X = \frac{\sin A \cos A + \sin B \cos B + \sin C \cos C}{\sin A \sin B \sin C}$

$\implies X = \frac{2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C}{2 \sin A \sin B \sin C}$

$\implies X = \frac{\sin 2 A + \sin 2 B + \sin 2 C}{2 \sin A \sin B \sin C} \to \left(1\right)$

Now , In a $\triangle A B C$, $A + B + C = \pi$

=>color(red)(A+B=pi-C and C=pi-(A+B)

So ,

$\diamond \sin 2 A + \sin 2 B + \sin 2 C \to A p p l y \left({F}_{1}\right)$

$= 2 \sin \left(\frac{2 A + 2 B}{2}\right) \cos \left(\frac{2 A - 2 B}{2}\right) + \sin 2 C$

$= 2 \sin \textcolor{red}{\left(A + B\right)} \cos \left(A - B\right) + \sin 2 C$

$= 2 \sin \textcolor{red}{\left(\pi - C\right)} \cos \left(A - B\right) + 2 \sin C \cos C$

$= 2 \sin C \cos \left(A - B\right) + 2 \sin C \cos C$

$= 2 \sin C \left[\cos \left(A - B\right) + \cos \textcolor{red}{C}\right]$

$= 2 \sin C \left[\cos \left(A - B\right) + \cos \textcolor{red}{\left(\pi - \left(A + B\right)\right)}\right]$

$= 2 \sin C \left[\cos \left(A - B\right) - \cos \left(A + B\right)\right] \to A p p l y \left({F}_{2}\right)$

=$2 \sin C \left[- 2 \sin \left(\frac{A - B + A + B}{2}\right) \sin \left(\frac{A - B - A - B}{2}\right)\right]$

$= 2 \sin C \left[- 2 \sin A \sin \left(- B\right)\right]$

$= 2 \sin C \left[2 \sin A \sin B\right]$

$= 4 \sin A \sin B \sin C$

$i . e .2 A + \sin 2 B + \sin 2 C = 4 \sin A \sin B \sin C \to \left(2\right)$

From $\left(1\right) \mathmr{and} \left(2\right)$

$X = \frac{4 \sin A \sin B \sin C}{2 \sin A \sin B \sin C}$

$\implies X = 2$

Jul 17, 2018

Please refer to another Proof in Explanation.

Explanation:

In $\Delta A B C , A + B + C = \pi$.

$\therefore B + C = \pi - A$.

$\therefore \cos \left(B + C\right) = \cos \left(\pi - A\right) = - \cos A$,

$i . e . , \cos B \cos C - \sin B \sin C = - \cos A$.

$\therefore \cos \frac{A}{\sin B \sin C} = \frac{\sin B \sin C - \cos B \cos C}{\sin B \sin C} ,$

$= \frac{\sin B \sin C}{\sin B \sin C} - \frac{\cos B \cos C}{\sin B \sin C}$.

$= 1 - \cot B \cot C$.

$\therefore \text{The Exp.} = 3 - \left\{\cot B \cot C + \cot C \cot A + \cot A \cot B\right\}$,

$= 3 - \left\{\frac{1}{\tan B \tan C} + \frac{1}{\tan C \tan A} + \frac{1}{\tan A \tan B}\right\}$,

$= 3 - \frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C} \ldots \ldots \ldots \ldots . . \left({\star}_{1}\right)$,

But, $B + C = \pi - A$.

$\therefore \tan \left(B + C\right) = \tan \left(\pi - A\right)$.

$\therefore \frac{\tan B + \tan C}{1 - \tan B \tan C} = - \tan A$.

$\therefore \tan B + \tan C = - \tan A + \tan A \tan B \tan C$.

$\therefore \tan A + \tan B + \tan C = \tan A \tan B \tan C \ldots \ldots \ldots \ldots \left({\star}_{2}\right)$.

Therefore, from $\left({\star}_{1}\right) \mathmr{and} \left({\star}_{2}\right)$

$\text{The Exp."=3-1=2," a prime} .$

Jul 17, 2018

Given

$A + B + C = \pi$

$\implies \cot \left(A + B\right) = \cot \left(\pi - C\right)$

$\implies \frac{\cot A \cot B - 1}{\cot B + \cot A} = - \cot C$

$\implies \cot A \cot B + \cot B \cot C + \cot C \cot A = 1$

Now 1st part of the given expression
$= \cos \frac{A}{\sin B \sin C}$

$= \cos \frac{\pi - \left(B + C\right)}{\sin B \sin C}$

$= \frac{- \cos B \cos C + \sin B \sin C}{\sin B \sin C}$

$= 1 - \cot B \cot C$

Similarly 2nd part

$= 1 - \cot A \cot B$

And 3rd part

$= 1 - \cot C \cot A$

So whole LHS

$= 3 - \left(\cot A \cot B + \cot B \cot C + \cot C \cot A\right)$

$= 3 - 1 = 2$,which is a prime