Đáp án:
$a + b + c = 11$
Giải thích các bước giải:
Ta có:
$\tan142^o30' = \tan142,5^o$
$= \tan(90^o + 52,5^o)$
$= -\cot52,5^o$
$= -\dfrac{1}{\tan(45^o + 7,5^o)}$
$= - \dfrac{1 - \tan7,5^o}{1 + \tan7,5^o}$
$= -\dfrac{\cos7,5^o - \sin7,5^o}{\cos7,5^o + \sin7,5^o}$
$= -\dfrac{(\cos7,5^o - \sin7,5^o)(\cos7,5^o - \sin7,5^o)}{(\cos7,5^o + \sin7,5^o)(\cos7,5^o - \sin7,5^o)}$
$= -\dfrac{(\cos7,5^o - \sin7,5^o)^2}{\cos^27,5^o - \sin^27,5^o}$
$= -\dfrac{1 - 2\sin7,5^o\cos7,5^o}{\cos15^o}$
$= -\dfrac{1 - \sin15^o}{\cos15^o}$
$= - \dfrac{1 - \sin(45^o - 30^o)}{\cos(45^o - 30^o)}$
$= - \dfrac{1 - \sin45^o\cos30^o + \sin30^o\cos45^o}{\cos45^o\cos30^o + \sin45^o\sin30^o}$
$= -\dfrac{1 - \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac{1}{2}\cdot\dfrac{\sqrt2}{2}}{\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}}$
$= -\dfrac{2\sqrt2 - \sqrt3 + 1}{\sqrt3 + 1}$
$= - \dfrac{(2\sqrt2 - \sqrt3 +1)(\sqrt3 -1)}{3 - 1}$
$= - \dfrac{2\sqrt6 - 2\sqrt2 - 3 + \sqrt3 + \sqrt3 -1}{2}$
$= 2 + \sqrt2- \sqrt3 - \sqrt6$
$\Rightarrow a + b + c = 2 + 3 + 6 = 11$
