Question

In$\Delta ABC$, right angled at B, if$\tan A=\dfrac{1}{\sqrt{3}}$then find the value of:
A. $\sin A\cos C+\cos A\sin C$
B. $\cos A\cos C-\sin A\sin C$

Answer 1

Hint: The value of$\tan A$is given in the question so we can find the value of angle A. From the value of $\tan A$i.e. $\tan A=\dfrac{1}{\sqrt{3}}$we can find the value of angle A i.e.$\angle A={{30}^{\circ }}$. It is given that$\angle B={{90}^{\circ }}$. And the sum of all the angles of the triangle is 180° so$\angle C={{60}^{\circ }}$. Now, we can easily write the values of:
$\sin A,\cos A,\sin C,\cos C$.

Complete step by step solution:
In$\Delta ABC$, it is given that:
$\tan A=\dfrac{1}{\sqrt{3}}$
The value of angle A when$\tan A=\dfrac{1}{\sqrt{3}}$is 30°.
In the below figure, we are showing a $\Delta ABC$ right angled at B.

Now,$\Delta ABC$is right angled at B so$\angle B={{90}^{\circ }}$.
Sum of the angles of a triangle = 180°
$\begin{align}
  & \angle A+\angle B+\angle C={{180}^{\circ }} \\
 & {{30}^{\circ }}+{{90}^{\circ }}+\angle C={{180}^{\circ }} \\
 & \angle C={{90}^{\circ }}-{{30}^{\circ }} \\
 & \angle C={{60}^{\circ }} \\
\end{align}$
The below figure is a right angled $\Delta ABC$ marked with angle $A={{30}^{0}},B={{90}^{0}},C={{60}^{0}}$.
 

From the above values of angles A and C we can find the value of$\sin A,\cos A,\sin C,\cos C$.
$\sin C=\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$
$\cos C=\cos {{60}^{\circ }}=\dfrac{1}{2}$
$\begin{align}
  & \sin A=\sin {{30}^{\circ }}=\dfrac{1}{2} \\
 & \cos A=\cos {{30}^{\circ }}=\dfrac{1}{2} \\
\end{align}$
Now, we are ready to solve the options given in the question.
A. $\sin A\cos C+\cos A\sin C$
Substituting the values of$\sin A,\cos A,\sin C,\cos C$in the above expression we get,
$\begin{align}
  & \dfrac{1}{2}\left( \dfrac{1}{2} \right)+\dfrac{\sqrt{3}}{2}\left( \dfrac{\sqrt{3}}{2} \right) \\
 & =\dfrac{1}{4}+\dfrac{3}{4} \\
 & =1 \\
\end{align}$
B. $\cos A\cos C-\sin A\sin C$
Substituting the values of$\sin A,\cos A,\sin C,\cos C$in the above expression we get,
$\begin{align}
  & \dfrac{\sqrt{3}}{2}\left( \dfrac{1}{2} \right)-\dfrac{1}{2}\left( \dfrac{\sqrt{3}}{2} \right) \\
 & =\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4} \\
 & =0 \\
\end{align}$

Note: The other way of solving the above problem is that:
If you carefully look at this expression $\sin A\cos C+\cos A\sin C$you will find this is the expansion of trigonometric identity$\sin \left( A+C \right)$. So, we can write this expression as$\sin \left( A+C \right)$.
From the sum of angles of a triangle we can write$A+C=\pi -B$so we can write$\sin \left( A+C \right)$as$\sin \left( \pi -B \right)$or$\sin B$where the value of$B={{90}^{0}}$.
So, the value of this expression$\sin A\cos C+\cos A\sin C$is $\sin {{90}^{0}}$(or 1).
Similarly, this expression$\cos A\cos C-\sin A\sin C$is the expansion of the identity$\cos \left( A+C \right)$.