Question

In a $△$ABC, right anged at B, if tan A = $\frac{1}{\sqrt{3}}$ , find the value of

1. sinA cosC + cosAsinC
2. cosAcosC – sinAsinC

​

Answer 1

We know that

tan A = $\frac{BC}{AB}=\frac{1}{\sqrt{3}}$

∴ BC : AB = 1 : $\sqrt{3}$

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

Then, AC = $\sqrt{A{B}^2+B{C}^2}$ ............ (Phythagoras theorm)

= $\sqrt{(\sqrt{3}k{)}^2+(k{)}^2}$ = $\sqrt{3k^2+k^2}$

= $\sqrt{4k^2}$ = 2k

Now, sin A = $\frac{BC}{AC}=\frac{k}{2k}=\frac{1}{2}$

cos A = $\frac{AB}{AC}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2}$

sin C = $\frac{AB}{AC}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2}$

and cos C = $\frac{BC}{AC}=\frac{BC}{AC}=\frac{k}{2k}=\frac{1}{2}$

(i) sin A cos C + cos A sin C = $\frac{1}{2}.\frac{1}{2}+\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}=1$

(ii) cos A cos C - sin A sin C = $\frac{\sqrt{3}}{2}.\frac{1}{2}-\frac{1}{2}.\frac{\sqrt{3}}{2}$ = 0