Question

In $△ABC$, right angled at $C$, if $\tan A=\frac{1}{\sqrt{3}}$, find the value of:
$\cos A\cos C-\sin A\sin C$

Answer 1

Give that $tanA=\sqrt{3},\angle B={90}^{\circ }$

$tanA=\sqrt{3}$

$\Rightarrow A=ta{n}^{-1}\sqrt{3}$

$A={60}^{\circ }$

We know that sum of all sides of a triangle are ${180}^{\circ }$

$\therefore \angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow {60}^{\circ }+{90}^{\circ }+\angle C={180}^{\circ }$

$\Rightarrow \angle C={180}^{\circ }-{150}^{\circ }$

$\Rightarrow \angle C={30}^{\circ }$

We know that $cos(A+C)=cosAcosC-sinAsinC$

then

$cos{60}^{\circ }cos{30}^{\circ }-sin{60}^{\circ }sin{30}^{\circ }=cos({60}^{\circ }+{30}^{\circ })$

$=cos{90}^{\circ }$

$=0$

$\therefore cosAcosC-sinAsinC=0$