Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(tan^2a+cot^2a=\left(tana+cota\right)^2-2=m^2-2\)
\(tan^4a+cot^4a=\left(tan^2a+cot^2a\right)^2-2=\left(m^2-2\right)^2-2\)
\(tan^6a+cot^6a=\left(tan^2a+cot^2a\right)^3-3\left(tan^2a+cot^2a\right)\)
\(=\left(m^2-2\right)^3-3\left(m^2-2\right)\)
\(m^2=\left(tana+cota\right)^2=\left(tana-cota\right)^2+4tana.cota\)
\(\Rightarrow m^2=\left(tana-cota\right)^2+4\ge4\)
\(\Rightarrow\left|m\right|\ge2\)
a) \(tan^2\alpha+cot^2\alpha=\left(tan\alpha+cot\alpha\right)^2-2tan\alpha cot\alpha\)
\(=m^2-2\).
b) \(tan^3\alpha+cot^3\alpha=\left(tan\alpha+cot\alpha\right)\)\(\left(tan^2\alpha-tan\alpha cot\alpha+cot^2\alpha\right)\)
\(=m\left(tan^2\alpha+cot^2\alpha-tan\alpha cot\alpha\right)\)
\(=m\left(m^2-2-2\right)=m\left(m^2-3\right)\).
\(\left(tana+cota\right)^2=m^2\)
\(\Leftrightarrow tan^2a+cot^2a+2=m^2\)
\(\Leftrightarrow tan^2a+cot^2a-2.tana.cota=m^2-4\)
\(\Leftrightarrow\left(tana-cota\right)^2=m^2-4\)
\(\Rightarrow tana-cota=\pm\sqrt{m^2-4}\)
\(0< a< \frac{\pi}{2}\Rightarrow sina;cosa;tana>0\)
\(tana+\frac{1}{tana}=3\Leftrightarrow tan^2a-3tana+1=0\) \(\Rightarrow\left[{}\begin{matrix}tana=\frac{3-\sqrt{5}}{2}\\tana=\frac{3+\sqrt{5}}{2}\end{matrix}\right.\)
- Với \(tana=\frac{3-\sqrt{5}}{2}\)
\(\Rightarrow cota=\frac{1}{tana}=\frac{3+\sqrt{5}}{2}\)
\(1+tan^2a=\frac{1}{cos^2a}\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}=\frac{2}{\sqrt{18-6\sqrt{5}}}\)
\(sina=\sqrt{1-cos^2a}=\frac{2}{\sqrt{18+6\sqrt{5}}}\)
\(cos\left(\frac{3\pi}{2}-a\right)=cos\left(2\pi-\frac{\pi}{2}-a\right)=-sina=...\)
\(sin\left(2\pi+a\right)=sina=...\)
\(tan\left(\pi-a\right)=-tana=...\)
\(cot\left(\pi+a\right)=cota=...\)
TH2: \(tana=\frac{3+\sqrt{5}}{2}\)
Tương tự như trên
a) \(sin6\alpha cot3\alpha cos6\alpha=2.sin3\alpha.cos3\alpha\dfrac{cos3\alpha}{sin3\alpha}-cos6\alpha\)
\(=2cos^23\alpha-\left(2cos^23\alpha-1\right)=1\) (Không phụ thuộc vào x).
b) \(\left[tan\left(90^o-\alpha\right)-cot\left(90^o+\alpha\right)\right]^2\)\(-\left[cot\left(180^o+\alpha\right)+cot\left(270^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+cot\left(90^o-\alpha\right)\right]^2\)\(-\left[cot\alpha+cot\left(90^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+tan\alpha\right]^2-\left[cot\alpha-tan\alpha\right]^2\)
\(=4tan\alpha cot\alpha=4\). (Không phụ thuộc vào \(\alpha\)).
\(tana-cota=3\Rightarrow\left(tana-cota\right)^2=9\)
\(\Rightarrow tan^2a+cot^2a-2=9\Rightarrow tan^2a+cot^2a=11\)
\(\frac{1}{tan^2a}+\frac{1}{cot^2a}=\frac{tan^2a+cot^2a}{\left(tana.cota\right)^2}=tan^2a+cot^2a=11\)
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 $\sin a=\cos a=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \sin a\cos a=\frac{1}{2}$
$E=\frac{\sin a.\cos a}{\tan ^2a+\cot ^2a}=\frac{\frac{1}{2}}{1+1}=\frac{1}{4}$
\(tan^3a+cot^3a=\left(tana+cota\right)^3-3tana.cota\left(tana+cota\right)\)
\(=m^3-3.1.m=m^3-3m\)
\(G=cot^2x-sin^2x.cot^2x+1-cot^2x=1-sin^2x.cot^2x\)
\(=1-sin^2x.\dfrac{cos^2x}{sin^2x}=1-cos^2x=sin^2x\)
2.
\(tana+cota=2\Rightarrow\left(tana+cota\right)^2=4\)
\(\Rightarrow tan^2a+cot^2a+2tana.cota=4\)
\(\Rightarrow tan^2a+cot^2a+2=4\)
\(\Rightarrow tan^2a+cot^2a=2\)
\(\left(tana+cota\right)^2=m^2\Leftrightarrow tan^2a+cot^2a+2=m^2\)
\(\Rightarrow tan^2a+cot^2a-2=m^2-4\)
\(\Leftrightarrow\left(tana-cota\right)^2=m^2-4\)
\(\Rightarrow\left|tana-cota\right|=\sqrt{m^2-4}\)
\(\Rightarrow tana-cota=\pm\sqrt{m^2-4}\)