Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Vì nên mặt khác suy ra .
Do đó .
b) Vì nên và .
c) Vì mặt khác nên .
Ta có .
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
Lời giải:
a.
$\tan a+\cot a=2\Leftrightarrow \tan a+\frac{1}{\tan a}=2$
$\Leftrightarrow \frac{\tan ^2a+1}{\tan a}=2$
$\Leftrightarrow \tan ^2a-2\tan a+1=0$
$\Leftrightarrow (\tan a-1)^2=0\Rightarrow \tan a=1$
$\cot a=\frac{1}{\tan a}=1$
$1=\tan a=\frac{\cos a}{\sin a}\Rightarrow \cos a=\sin a$
Mà $\cos ^2a+\sin ^2a=1$
$\Rightarrow \cos a=\sin a=\pm \frac{1}{\sqrt{2}}$
b.
Vì $\sin a=\cos a=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \sin a\cos a=\frac{1}{2}$
$E=\frac{\sin a.\cos a}{\tan ^2a+\cot ^2a}=\frac{\frac{1}{2}}{1+1}=\frac{1}{4}$
Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(tan\alpha< 0,cot\alpha< 0;cos\alpha< 0\).
Vì vậy: \(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{7}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{3}{4}:\dfrac{-\sqrt{7}}{4}=\dfrac{-3}{\sqrt{7}}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{-\sqrt{7}}{3}\).
\(A=\dfrac{2tan\alpha-3cot\alpha}{cos\alpha+tan\alpha}\)\(=\dfrac{2.\dfrac{-3}{\sqrt{7}}-3.\dfrac{-\sqrt{7}}{3}}{\dfrac{-\sqrt{7}}{4}+\dfrac{-3}{\sqrt{7}}}\)
\(=\dfrac{\dfrac{-6}{\sqrt{7}}+\sqrt{7}}{\dfrac{-7-12}{4\sqrt{7}}}\)\(=\dfrac{\dfrac{-6+7}{\sqrt{7}}.4\sqrt{7}}{-19}\)\(=\dfrac{\dfrac{1}{\sqrt{7}}.4\sqrt{7}}{-19}=-\dfrac{4}{19}\).
b) \(\dfrac{cos^2\alpha+cot^2\alpha}{tan\alpha-cot\alpha}=\dfrac{\left(-\dfrac{\sqrt{7}}{4}\right)^2+\left(\dfrac{-\sqrt{7}}{3}\right)^2}{\dfrac{-3}{\sqrt{7}}+\dfrac{\sqrt{7}}{3}}\)
\(=\dfrac{\dfrac{7}{16}+\dfrac{7}{9}}{\dfrac{-9+7}{3\sqrt{7}}}=\dfrac{\dfrac{175}{144}}{\dfrac{-2}{3\sqrt{7}}}=\dfrac{-175}{96\sqrt{7}}\).
a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(tan\alpha,cot\alpha>0\) và \(sin\alpha,cos\alpha< 0\).
\(\left\{{}\begin{matrix}tan\alpha-3cot\alpha=6\\tan\alpha cot\alpha=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=6+3cot\alpha\\\left(6+3cot\alpha\right)cot\alpha=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=6+3cot\alpha\\3cot^2\alpha+6cot\alpha-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=6+3cot\alpha\\cot\alpha=\dfrac{-3+2\sqrt{3}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}tan\alpha=3+2\sqrt{3}\\cot\alpha=\dfrac{-3+2\sqrt{3}}{3}\end{matrix}\right.\).
Có \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{1}{tan^2\alpha+1}\).
Có thể đề sai.
\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)
\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)
\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)
Lời giải:
$\frac{\pi}{2}< a< \pi$ nên $\sin a>0; \cos a< 0$
$-3=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=-3\cos a$
$\Rightarrow \sin ^2a=9\cos ^2a$
$\Rightarrow 10\sin ^2a=9(\sin ^2a+\cos ^2a)=9$
$\Rightarrow \sin ^2a=\frac{9}{10}$
$\Rightarrow \sin a=\frac{3}{\sqrt{10}}$
$\cos a=\frac{\sin a}{-3}=\frac{-1}{\sqrt{10}}$
$\cot a=\frac{1}{\tan a}=\frac{-1}{3}$
a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).
b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).
a) Ta có A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alphaA=tanα+tanα1tanα+3tanα1=tan2α+1tan2α+3=cos2α1cos2α1+2=1+2cos2α Suy ra A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}A=1+2⋅169=817.
b) B=\dfrac{\dfrac{\sin \alpha}{\cos ^{3} \alpha}-\dfrac{\cos \alpha}{\cos ^{3} \alpha}}{\dfrac{\sin ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{3 \cos ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{2 \sin \alpha}{\cos ^{3} \alpha}}=\dfrac{\tan \alpha\left(\tan ^{2} \alpha+1\right)-\left(\tan ^{2} \alpha+1\right)}{\tan ^{3} \alpha+3+2 \tan \alpha\left(\tan ^{2} \alpha+1\right)}B=cos3αsin3α+cos3α3cos3α+cos3α2sinαcos3αsinα−cos3αcosα=tan3α+3+2tanα(tan2α+1)tanα(tan2α+1)−(tan2α+1).
Suy ra B=\dfrac{\sqrt{2}(2+1)-(2+1)}{2 \sqrt{2}+3+2 \sqrt{2}(2+1)}=\dfrac{3(\sqrt{2}-1)}{3+8 \sqrt{2}}B=22+3+22(2+1)2(2+1)−(2+1)=3+823(2−1).