Question

In a right angle triangle $ABC$, right angle is at $B$, if $\tan A=\sqrt{3}$ then find the value of
$\sin A\cos C+\cos A\sin C$

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 $sin(A+C)=sinAcosC+cosAsinC$

then

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

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

$=1$

$\therefore sinAcosC+cosAsinC=1$