Tổng hợp câu hay và khó chương 6 - Phần 1

$\left\{ \begin{array}{l}\left( {a + 1} \right)\left( {b + 1} \right) = 2\\a = \dfrac{1}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = \dfrac{1}{3}\\a = \dfrac{1}{2}\end{array} \right.$

$\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x.\tan y}}$$= \dfrac{{\dfrac{1}{2} + \dfrac{1}{3}}}{{1 - \dfrac{1}{2}.\dfrac{1}{3}}} = 1$$\Rightarrow x + y = \dfrac{\pi }{4}$.

$C = \dfrac{1}{{\sin {{10}^o}}} + \dfrac{{\sqrt 3 }}{{\cos {{10}^o}}} = \dfrac{{\cos {{10}^{\rm{o}}} + \sqrt 3 \sin {{10}^{\rm{o}}}}}{{\sin {{10}^{\rm{o}}}\cos {{10}^{\rm{o}}}}} = \dfrac{{\dfrac{1}{2}\cos {{10}^{\rm{o}}} + \dfrac{{\sqrt 3 }}{2}\sin {{10}^{\rm{o}}}}}{{\dfrac{{2\sin {{10}^{\rm{o}}}\cos {{10}^{\rm{o}}}}}{4}}} = \dfrac{{4\sin {{40}^{\rm{o}}}}}{{\sin {{20}^{\rm{o}}}}} = 8\cos {20^{\rm{o}}}$.

${\sin ^2}x.{\tan ^2}x + 4{\sin ^2}x - {\tan ^2}x + 3{\cos ^2}x$ $= \left( {{{\sin }^2}x - 1} \right){\tan ^2}x + 4{\sin ^2}x + 3{\cos ^2}x$.

$=  - {\cos ^2}x.{\tan ^2}x + 4{\sin ^2}x + 3{\cos ^2}x$ $=  - {\sin ^2}x + 4{\sin ^2}x + 3\left( {1 - {{\sin }^2}x} \right) = 3$.

${\tan ^3}\alpha  + {\cot ^3}\alpha  = {\left( {\tan \alpha  + \cot \alpha } \right)^3} - 3\tan \alpha .\cot \alpha \left( {\tan \alpha  + \cot \alpha } \right) = {m^3} - 3m$.

$\sin \alpha .\cos \alpha  = \dfrac{1}{2}\left[ {{{\left( {\sin \alpha  + \cos \alpha } \right)}^2} - \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)} \right] = \dfrac{1}{2}\left[ {{{\left( {\dfrac{5}{4}} \right)}^2} - 1} \right] = \dfrac{9}{{32}}$.

Ta có:

$\cot \alpha  = 3 \Rightarrow \dfrac{{\cos \alpha }}{{\sin \alpha }} = 3 \Leftrightarrow \cos \alpha  = 3\sin \alpha$

Thay vào biểu thức đề bài ta được:

$\dfrac{{3\sin \alpha  - 2.3\sin \alpha }}{{12{{\sin }^3}\alpha  + 4.{{\left( {3\sin \alpha } \right)}^3}}}$$= \dfrac{{ - 3\sin \alpha }}{{120{{\sin }^3}\alpha }} =  - \dfrac{1}{{40}}.\dfrac{1}{{{{\sin }^2}\alpha }}$ $=  - \dfrac{1}{{40}}\left( {1 + {{\cot }^2}\alpha } \right)$ $=  - \dfrac{1}{{40}}\left( {1 + {3^2}} \right) =  - \dfrac{1}{4}$

${\left( {\dfrac{{\sin \alpha  + \tan \alpha }}{{\cos \alpha  + 1}}} \right)^2} + 1$$= {\left( {\dfrac{{\tan \alpha .\cos \alpha  + \tan \alpha }}{{\cos \alpha  + 1}}} \right)^2} + 1$$= {\left( {\tan \alpha } \right)^2} + 1$$= \dfrac{1}{{{{\cos }^2}\alpha }}$.

Ta có:

$\left( {1 + \tan a} \right)\left( {1 + \tan b} \right)$$= 1 + \tan a + \tan b + \tan a\tan b$ $= 1 + \tan \left( {a + b} \right)\left( {1 - \tan a\tan b} \right) + \tan a\tan b$

$= 1 + \tan \left( {{{20}^0} + {{25}^0}} \right)\left( {1 - \tan {{20}^0}.\tan {{25}^0}} \right) + \tan {20^0}.\tan {25^0}$ $= 1 + \tan {45^0}\left( {1 - \tan {{20}^0}\tan {{45}^0}} \right) + \tan {20^0}\tan {25^0}$ $= 1 + 1 - \tan {20^0}\tan {25^0} + \tan {20^0}\tan {25^0} = 2$

Ta có: ${\tan ^2}\dfrac{\alpha }{2} = \dfrac{{{{\sin }^2}\dfrac{\alpha }{2}}}{{{{\cos }^2}\dfrac{\alpha }{2}}} = \dfrac{{\dfrac{{1 - \cos \alpha }}{2}}}{{\dfrac{{1 + \cos \alpha }}{2}}} = \dfrac{{1 - \cos \alpha }}{{1 + \cos \alpha }}$ $\Leftrightarrow 1 - \cos \alpha  = {\tan ^2}\dfrac{\alpha }{2}\left( {1 + \cos \alpha } \right)$ 

Đặt $\tan \dfrac{\alpha }{2} = t$ thì $\cos \alpha  = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}$

Với $t = 2 \Rightarrow \cos \alpha  = \dfrac{{1 - 4}}{{1 + 4}} = \dfrac{{ - 3}}{5}$

Suy ra $B = \dfrac{{1 + 5\left( {\dfrac{{ - 3}}{5}} \right)}}{{3 - 2\left( {\dfrac{{ - 3}}{5}} \right)}} = \dfrac{{ - 2}}{{\dfrac{{21}}{5}}} = \dfrac{{ - 10}}{{21}}$.

Ta có: $0 < \alpha  < {90^0}$$\Leftrightarrow 0 < \dfrac{\alpha }{2} < {45^0}$$\Rightarrow 0 < \sin \dfrac{\alpha }{2} < \dfrac{{\sqrt 2 }}{2}$$\Leftrightarrow 0 < \sqrt {\dfrac{{x - 1}}{{2x}}}  < \dfrac{{\sqrt 2 }}{2}$$\Leftrightarrow x > 0$

${\sin ^2}\dfrac{\alpha }{2} + {\cos ^2}\dfrac{\alpha }{2} = 1$$\Rightarrow \cos \dfrac{\alpha }{2} = \sqrt {1 - {{\sin }^2}\dfrac{\alpha }{2}}$, vì $0 < \dfrac{\alpha }{2} < {45^0}$

$\Leftrightarrow \cos \dfrac{\alpha }{2} = \sqrt {\dfrac{{x + 1}}{{2x}}}$$\Rightarrow \tan \dfrac{\alpha }{2} = \sqrt {\dfrac{{x - 1}}{{x + 1}}}$

$tan\alpha  = \dfrac{{2\tan \dfrac{\alpha }{2}}}{{1 - {{\tan }^2}\dfrac{\alpha }{2}}} = \dfrac{{2\sqrt {\dfrac{{x - 1}}{{x + 1}}} }}{{1 - \dfrac{{x - 1}}{{x + 1}}}} = \sqrt {{x^2} - 1}$.

$A = {\tan ^2}\dfrac{\pi }{{24}} + {\cot ^2}\dfrac{\pi }{{24}}$

$= \dfrac{1}{{{{\cos }^2}\dfrac{\pi }{{24}}}} - 1 + \dfrac{1}{{{{\sin }^2}\dfrac{\pi }{{24}}}} - 1$

$= \dfrac{1}{{{{\cos }^2}\dfrac{\pi }{{24}}.\sin^2\dfrac{\pi }{{24}}}} - 2$

$= \dfrac{4}{{\sin^2\dfrac{\pi }{{12}}}} - 2$$= \dfrac{8}{{1 - \cos \dfrac{\pi }{6}}} - 2$

$\begin{array}{l} = \dfrac{8}{{1 - \frac{{\sqrt 3 }}{2}}} - 2 = \dfrac{{16}}{{2 - \sqrt 3 }} - 2\\ = \dfrac{{16 - 4 + 2\sqrt 3 }}{{2 - \sqrt 3 }} = \dfrac{{12 + 2\sqrt 3 }}{{2 - \sqrt 3 }} \end{array}$

Vì $0 < x < \dfrac{\pi }{2}$ nên $\cos \dfrac{x}{n} > 0$, $\forall n \in {\mathbb{N}^*}$

$\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos x} } }$$= \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos \dfrac{x}{2}} }$$= \sqrt {\dfrac{1}{2} + \dfrac{1}{2}\cos \dfrac{x}{4}}  = \cos \dfrac{x}{8}$

Vậy $n = 8$.