Question

sa se arate ca {( sin α+ cos β)^2 + ( sin α - cos β)^2) } / 2 ctg α = tg α

Answer 1

$(\sin{\alpha}+\cos{\alpha})^{2}=\sin{\alpha}^{2}+\cos{\alpha}^{2}+2\sin{\alpha}\cos{\alpha}=1+2\sin{\alpha}\cos{\alpha}$
$(\sin{\alpha}-\cos{\alpha})^{2}=\sin{\alpha}^{2}+\cos{\alpha}^{2}-2\sin{\alpha}\cos{\alpha}=1-2\sin{\alpha}\cos{\alpha}<span>$
$\frac{(\sin{\alpha}+\cos{\alpha})^{2}+(\sin{\alpha}-\cos{\alpha})^{2}}{2ctg\alpha}=\frac{1+2\sin{\alpha}\cos{\alpha}+1-2\sin{\alpha}\cos{\alpha}}{<span>2ctg\alpha</span>}=\frac{2}{2ctg\alpha}=\frac{1}{ctg\alpha}=tg\alpha$