Ta có: $\alpha + \beta = 90^o$
$\Rightarrow \begin{cases} sin\alpha = cos\beta\\cos\alpha = sin\beta\\tan\alpha = cot\beta\\cot\alpha = tan\beta\end{cases}$
Ta được:
$2cot34^o.cot56^o + sin^229^o - 3.\dfrac{tan52^o}{cot38^o} + sin^261^o$
$= 2.tan56^o.cot56^o + sin^229^o - 3.\dfrac{cot38^o}{cot38^o} + cos^229^o$
$= 2.1 + 1 - 3.1 = 0$