In triangle $ABC$, right-angled at $B$, if $tan\ A = \frac{1}{\sqrt3}$, find the value of:
(i) $sin\ A\ cos\ C + cos\ A\ sin\ C$
(ii) $cos\ A\ cos\ C - sin\ A\ sin\ C$

To do:

We have to find the value of

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

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

Solution:  

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

This implies,

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

We know that,

In a right-angled triangle $ABC$ with a right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ A=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ A=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$sin\ C=\frac{Opposite}{Adjacent}=\frac{AB}{AC}$

$cos\ C=\frac{Adjacent}{Opposite}=\frac{BC}{AC}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(1)^2+(\sqrt3)^2$

$\Rightarrow AC^2=1+3$

$\Rightarrow AC=\sqrt{4}=2$

Therefore,

$sin\ A=\frac{BC}{AC}=\frac{1}{2}$

$cos\ A=\frac{AB}{AC}=\frac{\sqrt3}{2}$

$sin\ C=\frac{AB}{AC}=\frac{\sqrt3}{2}$

$cos\ C=\frac{BC}{AC}=\frac{1}{2}$

This implies,

(i) $\sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C}=\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$

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

$=\frac{4}{4}$

$=1$

(ii) $\cos \mathrm{A} \cos \mathrm{C}-\sin \mathrm{A} \sin \mathrm{C}=\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}$

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

$=0$

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Updated on: 10-Oct-2022

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