In triangle ABC, right-angled at B, if tan A = \(\dfrac{1}{\sqrt{3}}\), find:
(i) sin A cos C + cos A sin C
(ii) cos A cos C − sin A sin C
(i) 1
(ii) 0
Given: In right triangle \(\triangle ABC\), \(\angle B = 90^\circ\) and \(\tan A = \dfrac{1}{\sqrt{3}}\).
Step 1: Convert \(\tan A\) into side ratio
In a right-angled triangle, \(\tan A = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{BC}{AB}\).
So, \(\dfrac{BC}{AB} = \dfrac{1}{\sqrt{3}}\). Take \(BC = 1\) and \(AB = \sqrt{3}\) (any proportional sides work).
Step 2: Find the hypotenuse using Pythagoras theorem
\(AC^2 = AB^2 + BC^2\)
\(AC^2 = (\sqrt{3})^2 + 1^2 = 3 + 1 = 4\)
\(\Rightarrow AC = 2\).
Step 3: Write \(\sin\) and \(\cos\) using side ratios
For angle \(A\):
\(\sin A = \dfrac{BC}{AC} = \dfrac{1}{2}\), \(\cos A = \dfrac{AB}{AC} = \dfrac{\sqrt{3}}{2}\).
For angle \(C\): (opposite to \(C\) is \(AB\), adjacent to \(C\) is \(BC\))
\(\sin C = \dfrac{AB}{AC} = \dfrac{\sqrt{3}}{2}\), \(\cos C = \dfrac{BC}{AC} = \dfrac{1}{2}\).
Step 4: Substitute to find the required values
(i) \(\sin A\cos C + \cos A\sin C\)
= \(\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right) + \left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)\)
= \(\dfrac{1}{4} + \dfrac{3}{4} = 1\).
So, (i) = 1.
(ii) \(\cos A\cos C - \sin A\sin C\)
= \(\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{1}{2}\right) - \left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)\)
= \(\dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4} = 0\).
So, (ii) = 0.