Question

In a $△$ABC, right angles at C, if $tanA=\frac{1}{\sqrt{3}}$, then find the value of $sinAsinB+cosAcosB$.

Answer 1

First we need to determine the values of angles A and B and determine the value of their sine and cosine.
We are given that $tanA=\frac{1}{\sqrt{3}}$. there $\angle$ A=30º
$\angle A+\angle B+\angle C$ = 180º
therefore as $\angle$C is 90º, we get $\angle$B= 60º

sin A = sin 30º = $\frac{1}{2}$

sin B = sin 60º = $\frac{\sqrt{3}}{2}$

cos A = cos 30º = $\frac{\sqrt{3}}{2}$

cos B = cos 60º = $\frac{1}{2}$

For the final step, we need to substitute the corresponding values in the given expression

On substituting the respective values, we get
=$\frac{1}{2}\times \frac{\sqrt{3}}{2}$+$\frac{\sqrt{3}}{2}\times \frac{1}{2}$
=$\frac{2\left(\sqrt{3}\right)}{4}$
=$\frac{\sqrt{3}}{2}$= answer