In the adjoining figure, the value of sin B cos C + sin C cos B is

1. 0

2. 1

3. $\dfrac{5}{3}$

4. 2

In right angle triangle ABC,

⇒ BC² = AB² + AC²

⇒ 10² = (6)² + (AC)²

⇒ AC² = 10² - 6²

⇒ AC² = 100 - 36

⇒ AC² = 64

⇒ AC = $\sqrt{64}$ = 8 cm.

By formula,

sin B = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{AC}{BC} = \dfrac{8}{10}$.

sin C = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{AB}{BC} = \dfrac{6}{10}$.

cos B = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

= $\dfrac{AB}{BC} = \dfrac{6}{10}$.

cos C = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

= $\dfrac{AC}{BC} = \dfrac{8}{10}$.

Substituting values in sin B cos C + sin C cos B we get :

$\Rightarrow \text{sin B cos C + sin C cos B} = \dfrac{8}{10} \times \dfrac{8}{10} + \dfrac{6}{10} \times \dfrac{6}{10} \\[1em] = \dfrac{64}{100} + \dfrac{36}{100} \\[1em] = \dfrac{100}{100} = 1.$

Hence, Option 2 is the correct option.