Question

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

Answer 1

Hint:We know that the given triangle ABC is a right angled triangle and that the sum of the three angles of the same is equal to $180{}^\circ$. So, to solve this question, we will obtain the values of angle A and angle C and put the obtained values of angle A and angle C in the given trigonometric expression to get the desired answer.

Complete step-by-step answer:
It has been given in the question that ABC is a right angled triangle and that angle C is equal to angle A. It is also given that $\angle B=90{}^\circ$. And we have been asked to find the value of the given trigonometric expression, $\sin A\sin B+\cos A\cos B$. Now, we know that the sum of all the three angles of a triangle is equal to $180{}^\circ$. So, it means that in triangle ABC, we have,
$\angle A+\angle B+\angle C=180{}^\circ \ldots \ldots \ldots \left( i \right)$
We have also been given in the question that, $\angle B=90{}^\circ$. So, we can represent the triangle ABC as in the figure given below.

So, we will put the value of angle B, that is, $\angle B=90{}^\circ$ in equation (i). So, we get the equation as follows,
$\begin{aligned} & \angle A+\angle C+90{}^\circ =180{}^\circ \\\ & \Rightarrow \angle A+\angle C=180{}^\circ -90{}^\circ \\\ & \Rightarrow \angle A+\angle C=90{}^\circ \ldots \ldots \ldots \left( ii \right) \\\ \end{aligned}$
Now, we know that, $\angle A=\angle C$, is given in the question. So, we put that in equation (ii). So, we get,
$\begin{aligned} & \angle C+\angle C=90{}^\circ \\\ & \Rightarrow 2\angle C=90{}^\circ \\\ & \Rightarrow \angle C=\dfrac{90{}^\circ }{2} \\\ & \Rightarrow \angle C=45{}^\circ \\\ \end{aligned}$
Which means that, $\angle A=45{}^\circ$also. Now, we will put the obtained values of $\angle A=45{}^\circ$ and the given value of $\angle B=90{}^\circ$ in the given trigonometric expression,
$\begin{aligned} & \sin A\sin B+\cos A\cos B \\\ & \Rightarrow \sin 45{}^\circ \sin 90{}^\circ +\cos 45{}^\circ \cos 90{}^\circ \\\ \end{aligned}$
We know that the values of $\sin 45{}^\circ =\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\sin 90{}^\circ =1$ and $\cos 90{}^\circ =0$. So, by putting these values in the above trigonometric expression, we get,
$\begin{aligned} & \dfrac{1}{\sqrt{2}}\times 1+\dfrac{1}{\sqrt{2}}\times 0 \\\ & \Rightarrow \dfrac{1}{\sqrt{2}}+0=\dfrac{1}{\sqrt{2}} \\\ \end{aligned}$
Therefore, the value of the given expression, $\sin A\sin B+\cos A\cos B$ is $\dfrac{1}{\sqrt{2}}$.

Note: There is a possibility of the students getting confused with the values of $\cos 90{}^\circ$ and $\sin 90{}^\circ$ as 1 and 0 respectively, which would be incorrect. The students will get the same answer of $\dfrac{1}{\sqrt{2}}$, but it will be incorrect conceptually. So, it is advisable that the students memorise the values of the trigonometric standard angles.