In a triangle ABC , A= B+ C then what is ((B)/(2))+ ((B)/(2))equal...

Given $\sin A = \cos B + \cos C$

Using the half-angle and sum-to-product formulas:

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

In $\Delta ABC$, $A + B + C = \pi$, so $\dfrac{B+C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$.

This yields $\cos\left(\dfrac{B+C}{2}\right) = \sin\left(\dfrac{A}{2}\right)$.

Substituting this into the equation:

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

Since $A \in (0, \pi)$, $\sin\left(\dfrac{A}{2}\right) \neq 0$. Dividing both sides by $2 \sin\left(\dfrac{A}{2}\right)$:

$\cos\left(\dfrac{A}{2}\right) = \cos\left(\dfrac{B-C}{2}\right)$

Since both $\dfrac{A}{2} and$\dfrac{B-C}{2}$lie in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, we have:

$\dfrac{A}{2} = \pm \dfrac{B-C}{2}$

This implies either $A = B - C$ or $A = C - B$.

If $A = B - C$, then $B = A + C$. Since $A + B + C = \pi$, we get $2B = \pi \Rightarrow B = \dfrac{\pi}{2}$.

If $A = C - B$, then $C = A + B$. Since $A + B + C = \pi$, we get $2C = \pi \Rightarrow C = \dfrac{\pi}{2}$.

Simplifying the required expression:

$\tan\left(\dfrac{B}{2}\right) + \cot\left(\dfrac{B}{2}\right) = \dfrac{\sin\left(\dfrac{B}{2}\right)}{\cos\left(\dfrac{B}{2}\right)} + \dfrac{\cos\left(\dfrac{B}{2}\right)}{\sin\left(\dfrac{B}{2}\right)} = \dfrac{\sin^2\left(\dfrac{B}{2}\right) + \cos^2\left(\dfrac{B}{2}\right)}{\sin\left(\dfrac{B}{2}\right) \cos\left(\dfrac{B}{2}\right)} = \dfrac{2}{\sin B}$

If $B = \dfrac{\pi}{2}$, the value is $\dfrac{2}{\sin\left(\dfrac{\pi}{2}\right)} = 2$.

If $C = \dfrac{\pi}{2}$, then $B \in \left(0, \dfrac{\pi}{2}\right)$, which means $\sin B < 1$ and $\dfrac{2}{\sin B} > 2$.

Since the given options are constants and the minimum possible value is $2$, the only valid choice among the options is $2$.

Answer: $2$