Question

In triangle ABC, right-angled at B, if tan A = $1/\surd 3$, then, cos Acos C - sinAsin C =___.

Answer 1

Let $\Delta$ABC in which $\angle B={90}^{\circ }$,
According to 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
$sinA=\frac{BC}{AC}=\frac{1}{2}$
$sinC=\frac{AB}{AC}=\frac{\sqrt{3}}{2}$
$cosA=\frac{AB}{AC}=\frac{\sqrt{3}}{2}$
$cosC=\frac{BC}{AC}=\frac{1}{2}$
$cosAcosC-sinAsinC=(\frac{\sqrt{3}}{2}\times \frac{1}{2})-(\frac{1}{2}\times \frac{\sqrt{3}}{2})=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$