Question

In triangle ABC, right-angled at B, if tan A =$\frac{ 1}{\sqrt{3}}$. find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C

Answer 1

tan A = $\frac{1}{\sqrt3}$

$\frac{BC}{AB} = \frac{1}{\sqrt3}$

If BC is k, then AB will be $\sqrt3$ k, , where k is a positive integer.
In $Δ$ ABC,
$\text{AC}^ 2 =\text{ AB}^ 2 + \text{BC}^ 2$
$\text{AC}2=(\sqrt3 k)^2+(k)^2$
AC2=3k2+k2
AC2=4k2
AC = 2k

$\text{Sin A} =\frac{ BC}{AC} =\frac{ 1}{2}$

$\text{Cos A} =\frac{ AB}{AC} = \frac{\sqrt3}{2}$

$\text{Sin C} =\frac{ AB}{AC} = \frac{\sqrt3}{2}$

$\text{Cos C} = \frac{BC}{AC} = \frac{1}{2}$

(i) sin A cos C + cos A sin C =$(\frac{1}{2}) ×(\frac{1}{2} )+\frac{ \sqrt3}{2} ×\frac{\sqrt3}{2} = \frac{1}{4} +\frac{ 3}{4} = 1$

(ii) cos A cos C – sin A sin C =$(\frac{\sqrt3}{2} )(\frac{1}{2}) – (\frac{1}{2}) (\frac{\sqrt3}{2} ) = 0$