In a $ \Delta A B C $ right angled at $ B, \angle A=\angle C $. Find the values of
$ \sin A \sin B+\cos A \cos B $

 Given:

In a $\triangle ABC$ right angled at B, $\angle A = \angle C$.

To do:

We have to find the value of $\sin\ A\ sin\ B + \cos\ A\ cos\ B$.

Solution:  

We know that sum of the angles in a triangle is $180^{\circ}$.

This implies,

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

$\angle A+90^{\circ}+\angle A=180^{\circ}$             ($\angle B=90^{\circ}$)

$2\angle A=180^{\circ}-90^{\circ}$

$\angle A=\frac{90^{\circ}}{2}$

$\angle A=\angle C=45^{\circ}$

Therefore,

$\sin\ A\ sin\ B + \cos\ A\ cos\ B=\sin\ 45^{\circ}\ sin\ 90^{\circ} + \cos\ 45^{\circ}\ cos\ 90^{\circ}$$

$=\frac{1}{\sqrt2}\times1+\frac{1}{\sqrt2}\times0$          (Since $\sin 45^{\circ}=\frac{1}{\sqrt2}, \sin 90^{\circ}=1, \cos 90^{\circ}=0$)       

$=\frac{1}{\sqrt2}+0$

$=\frac{1}{\sqrt2}$

The value of $\sin\ A\ sin\ B + \cos\ A\ cos\ B$ is $\frac{1}{\sqrt2}$. 

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Updated on: 10-Oct-2022

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