Ta có:
$\tan13^o=\cot(90^o-13^o)=\cot77^o$
$\cos^226^o=\sin^2(90^o-26^o)=\sin^264^o$
$\cos^271^o=\sin^2(90^o-71^o)=\sin^219^o$
$\sin^234^o=\cos^2(90^o-34^o)=\cos^256^o$
$\sin^215^o=\cos^2(90^o-15^o)=\cos^275^o$
$\forall \alpha, \sin^2\alpha+\cos^2\alpha=1$
Vậy:
$A=\dfrac{3}{2}-\dfrac{1-1}{1+1}$
$=\dfrac{3}{2}$