In triangle ABC, right-angled at B, if tan A = $\dfrac{1}{\sqrt{3}}$, find the value of :

(i) sin A cos C + cos A sin C

(ii) cos A cos C - sin A sin C

Let us consider a right angle triangle ABC.

Given,

tan A = $\dfrac{1}{\sqrt{3}}$

We know that,

tan A = $\dfrac{\text{Side opposite to ∠A}}{\text{Side adjacent to ∠A}}$

Substituting values, we get :

$\dfrac{1}{\sqrt{3}} = \dfrac{BC}{AB}$

Let AB = $\sqrt{3}$k and BC = k.

In △ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = $(\sqrt{3}k)^2 + k^2$

⇒ AC² = 3k² + k²

⇒ AC² = 4k²

⇒ AC = $\sqrt{4k^2}$ = 2k.

We know that,

sin A = $\dfrac{\text{Side opposite to ∠A}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{k}{2k} = \dfrac{1}{2}$

cos A = $\dfrac{\text{Side adjacent to ∠A}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2}$

sin C = $\dfrac{\text{Side opposite to ∠C}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2}$

cos C = $\dfrac{\text{Side adjacent to ∠C}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{k}{2k} = \dfrac{1}{2}$

(i) Substituting values of sin A, cos C, sin C and cos A in sin A cos C + cos A sin C, we get :

$\Rightarrow \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{1}{4} + \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{4}{4} \\[1em] \Rightarrow 1.$

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

(ii) Substituting values of cos A, cos C, sin A and sin C in cos A cos C - sin A sin C, we get :

$\Rightarrow \dfrac{\sqrt{3}}{2} \times \dfrac{1}{2} - \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4} \\[1em] \Rightarrow 0.$

Hence, cos A cos C - sin A sin C = 0.