In $$ riangle ABC$$, if $$\sin A\cos(\dfrac{\pi}{2}-B)=1-\sin (\dfrac{\pi}{2}-A)\cos B $$ , then the triangle is a/n

Question

In  $$	riangle  ABC$$,   if $$\sin A\cos(\dfrac{\pi}{2}-B)=1-\sin (\dfrac{\pi}{2}-A)\cos B $$  , then the triangle is a/n

Uniteasy Admin · Accepted Answer

$$\sin A\cos(\dfrac{\pi}{2}-B)+\sin (\dfrac{\pi}{2}-A)\cos B $$ = 1

$$\sin A(\cos\dfrac{\pi}{2} \cos B +\sin \dfrac{\pi}{2}  \sin B)+ \cos B(\sin \dfrac{\pi}{2}   \cos A- \cos \dfrac{\pi}{2}  \sin A) = 1 $$
$$ \sin A \sin B + \cos A \cos B = 1$$
$$\cos(A-B) = 1$$
$$A-B =0  $$

Therefore, the triangle is an isosceles  triangle

Multiple Choice Question (MCQ)

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