Question

Question 9 (i)
In triangle ABC, right-angled at B, if $tanA=\frac{1}{\sqrt{3}}$, find the value of:
(i) sinAcosC + cosAsinC

Answer 1

Let $\Delta$ABC in which $\angle B={90}^{\circ }$,
According to the question,

Tan A = $\frac{BC}{AB}$ = $\frac{1}{\sqrt{3}}$
Let AB = $\sqrt{3}k$ and BC = k, where k is a positive real number.
By Pythagoras theorem in $\Delta$ABC; we get,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=(\sqrt{3}k{)}^2+(k{)}^2$
$A{C}^2=3k^2+k^2$
$A{C}^2=4k^2$
AC = 2k
sin A = $\frac{BC}{AC}$ = $\frac{1}{2}$
sin C = $\frac{AB}{AC}$ = $\frac{\sqrt{3}}{2}$
cos A = $\frac{AB}{AC}$ = $\frac{\sqrt{3}}{2}$
cos C = $\frac{BC}{AC}$ = $\frac{1}{2}$
(i) $sinAcosC+cosAsinC=(\frac{1}{2}\times \frac{1}{2})+(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2})=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1$