$\triangle$ $ABC ,$ માં $\angle B$ કાટખૂણો છે. જો $\tan A =\frac{1}{\sqrt{3}},$ હોય, તો નિમ્નલિખિત મૂલ્ય શોધો.

$(i)$ $\sin A \cos C+\cos A \sin C$

$(ii)$ $\cos A \cos C-\sin A \sin C$

$\tan A =\frac{1}{\sqrt{3}}$

$\frac{ BC }{ AB }=\frac{1}{\sqrt{3}}$

If $B C$ is $k$, then $A B$ will be $\sqrt{3} k,$ where $k$ is a positive integer.

$\ln \triangle ABC ,$

$-A C^{2}=A B^{2}+B C^{2}$

$(\sqrt{3} k)^{2}+(k)^{2}$

$=3 k^{2}+k^{2}=4 k^{2}$

$AC =2 k$

$\sin A=\frac{\text { Side opposite to } \angle A }{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{k}{2 k}=\frac{1}{2}$

$\cos A=\frac{\text { Side adjacent to } \angle A }{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}$

$\sin C=\frac{\text { Side opposite to } \angle C }{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}$

$\cos C=\frac{\text { Side adjacent to } \angle C }{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{k}{2 k}=\frac{1}{2}$

$(i)$ $\sin A \cos C+\cos A \sin C$

$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{4}+\frac{3}{4}$

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

$(ii)$ $\cos A \cos C-\sin A \sin C$

$=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$