ข้อ25
$sec^2A+cot^2A=8$
$sec^2A-tan^2A=1$
ลบกัน$\frac{1}{tan^2A} +tan^2A=7$
ให้$x=tan^2A $ได้
$x^2+\frac{1}{x^2} =7$
$x^2+\frac{1}{x^2}+2 =9$
$x+\frac{1}{x} =3$
$\frac{sinA}{cosA} +\frac{cosA}{sinA} =\frac{1}{sinAcosA} =3$
$sinAcosA=\frac{1}{3} $
$sin^2+2sincos+cos^2=(sin+cos)^2$
$=\sqrt{1+\frac{2}{3} } =\sqrt{\frac{10}{6} }$
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