From the given figure, write down the values of :

(i) sin B

(ii) tan B

(iii) cos C

(iv) cot C

(v) (sin B cos C + cos B sin C)

(vi) (sec² C - tan² C)

Given a right triangle ABC with hypotenuse BC = 17 units and AB = 15 units.

First, find AC using the Pythagoras theorem :

BC² = AB² + AC²

AC² = BC² - AB²

AC² = 17² - 15²

AC² = 289 - 225

AC² = 64

AC = $\sqrt{64}$

AC = 8 units

(i) sin B = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AC}{BC} = \dfrac{8}{17}$

(ii) tan B = $\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AC}{AB} = \dfrac{8}{15}$

(iii) cos C = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{AC}{BC} = \dfrac{8}{17}$

(iv) cot C = $\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{AC}{AB} = \dfrac{8}{15}$

(v) We have to find

sin B cos C + cos B sin C

First we will find the values of cos B & sin C

cos B = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{AB}{BC} = \dfrac{15}{17}$

sin C = $\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{AB}{BC} = \dfrac{15}{17}$

Substituting the values, we get :

$\text{sin B cos C + cos B sin C} = \dfrac{8}{17}\times \dfrac{8}{17} + \dfrac{15}{17} \times \dfrac{15}{17} \\[1em] = \dfrac{64}{289} + \dfrac{225}{289} \\[1em] = \dfrac{289}{289} \\[1em] = 1.$

Hence, sin B cos C + cos B sin C = 1.

(vi) We have to find out

sec²C - tan²C

First we will find out the values of sec C & tan C

sec C = $\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{BC}{AC} = \dfrac{17}{8}$

tan C = $\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{AB}{AC} = \dfrac{15}{8}$

Now putting the values of sec C & tan C

sec²C - tan²C

= $\Big(\dfrac{17}{8}\Big)^2 - \Big(\dfrac{15}{8}\Big)^2$

= $\dfrac{289}{64} - \dfrac{225}{64} = \dfrac{289 - 225}{64}$

= $\dfrac{64}{64}$

= 1.

Hence, sec²C - tan²C = 1.