In a triangle ABC that is right angled at C, ∠A = ∠B. The value of sinA sinB + cosA cosB is:

Correct option is C

Given:

In a triangle ABC that is right angled at C, ∠A = ∠B

Formula Used:

Sum of angles of triangle = $$180^o$$​​

​$$\sin 45^o = \frac{1}{\sqrt{2}}$$​​

​$$\cos 45^o = \frac{1}{\sqrt{2}}$$​​

Solution:

If $$\angle A = \angle B\ and\ \angle C = 90^o$$​​

Then $$\angle A + \angle B +\angle C = 180^ o$$​​

​$$2 \angle A =180^o - 90^o$$​​

​$$\angle A = \angle B = 45^o$$​​

Then $$\sin A\sin B + \cos A\ \cos B :$$​​

​$$\sin45^o \sin45^o + \cos45^o \cos45^o$$​

​$$\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)$$​​

​$$\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)$$​​

= 1

Alternative Method:

We know that $$\angle A = \angle B = 45^o$$​​

And $$\cos(A-B) =sinA sinB + cosA cosB$$​​

Hence $$\cos(45^o - 45^o) =\cos(0^o) = 1$$​​