Tam giác ABC vuông tại A
=> góc B + góc C = 90o
=> sin C = cos B = \(\dfrac{1}{2}\)
Có sin2B + cos2B = 1
=> sin2B + \(\dfrac{1}{2}\) = 1
=> sin2B = \(\dfrac{1}{2}\)
=> sin B = \(\sqrt{\dfrac{1}{2}}\)
Ta có : tan B = \(\dfrac{\sqrt{\dfrac{1}{2}}}{\dfrac{1}{2}}\)
cot B = \(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{1}{2}}}\)