toán 9

T

thaotran19

a)$1+tan^2 \alpha = 1+ \dfrac{sin^2 \alpha}{cos^2 \alpha}$
$=1+ \dfrac{1-cos^2 \alpha}{cos^2 \alpha}$
$=1+\dfrac{1}{cos^2 \alpha}-1$
$=\dfrac{1}{cos^2 \alpha}$( đpcm) :D
 
Top Bottom