Question

In $△$ABC, $\angle B={90}^{\circ }$
If $tanA=\frac{1}{\sqrt{3}}$, then
$sinA.cosC+cosA.sinC$ and and, $cosA.cosC-sinA.sinC$ is equal to

• A. 0 and 1 respectively
• B. 1 and 0 respectively
• C. 1 and 2 respectively
• D. 2 and 1 respectively

Answer 1

The correct option is B 1 and 0 respectively

Given $tanA=\frac{1}{\sqrt{3}}$, and $\angle B={90}^{\circ }$ in $△$ABC.

We know that $tan{30}^{\circ }=\frac{1}{\sqrt{3}}$

$\therefore \angle A={30}^{\circ }$

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

$\Rightarrow \angle C={180}^{\circ }-\angle B-\angle A={180}^{\circ }-{30}^{\circ }-{90}^{\circ }={60}^{\circ }$

Now, $sinA=sin{30}^{\circ }=\frac{1}{2}$

$cosA=cos{30}^{\circ }=\frac{\sqrt{3}}{2}$

$sinC=sin{60}^{\circ }=\frac{\sqrt{3}}{2}$

$cosC=cos{60}^{\circ }=\frac{1}{2}$

The first expression:
sin A cos C + cos A sin C
$=\frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}$

$=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1$

The second expression,
cos A cos C - sin A sin C
$=\frac{\sqrt{3}}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{1}{2}$

$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$