[Toán 10] $\tan A = - \tan(B + C)$

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noinhobinhyen

Trang tam giác ABC thì $A+B+C = 180 ; \dfrac{A}{2}+\dfrac{B}{2}+\dfrac{C}{2}=90^o$

+$\sinA = \sin(180^o-A)=\sin(B+C)$

+$\cos\dfrac{A}{2} = \sin(90^o-\dfrac{A}{2})=\sin(\dfrac{B}{2}+\dfrac{C}{2})$

+$\tanA = -\tan(180^o-A)=-\tan(B+C)$
 
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