Multiple Choice Question (MCQ)
In \triangle ABC, if \sin A\cos(\dfrac{\pi}{2}-B)=1-\sin (\dfrac{\pi}{2}-A)\cos B , then the triangle is a/n
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×
acute triangle
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×
right triangle
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×
obtuse triangle
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✓
isosceles triangle
Answer
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\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