CM \(1+tg^2\alpha=\frac{1}{cos^2\alpha}\)
\(1+cotg^2\alpha=\frac{1}{sin^2\alpha}\)
Những câu hỏi liên quan
cm các đẳng thức:
a) \(\frac{1+\sin^2\alpha}{1-\sin^2\alpha}=1+2\tan^2\alpha\)
b) \(\frac{\cos\alpha}{1+\sin\alpha}+\tan\alpha=\frac{1}{\cos\alpha}\)
c) \(\frac{\sin\alpha}{1+\cos\alpha}+\frac{1+\cos\alpha}{\sin\alpha}=\frac{2}{\sin\alpha}\)
\(\frac{1+sin^2a}{1-sin^2a}=\frac{1+sin^2a}{cos^2a}=\frac{1}{cos^2a}+\frac{sin^2a}{cos^2a}=1+tan^2a+tan^2a=1+2tan^2a\)
\(\frac{cosa}{1+sina}+tana=\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina+sin^2a}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sina}{1+cosa}+\frac{1+cosa}{sina}=\frac{sin^2a+cos^2a+2cosa+1}{\left(1+cosa\right)sina}=\frac{2+2cosa}{\left(1+cosa\right)sina}=\frac{2\left(1+cosa\right)}{\left(1+cosa\right)sina}=\frac{2}{sina}\)
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chứng minh với góc nhọn \(\alpha\) túy ý có;
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
cotg\(\alpha\)=\(\frac{\cos\alpha}{sin\alpha}\)
\(\tan\alpha\) . cotg \(\alpha\)=1
\(\sin^2\alpha+\cos^2\alpha=1\)
a/ \(\sin\alpha=\frac{C_đ}{C_h}\)
\(\cos\alpha=\frac{C_k}{C_h}\)
\(\Rightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{C_đ}{C_h}}{\frac{C_k}{C_h}}=\frac{C_đ}{C_k}=\tan\alpha\)
b/ \(\frac{\cos\alpha}{\sin\alpha}=\frac{\frac{C_k}{C_h}}{\frac{C_đ}{C_h}}=\frac{C_k}{C_đ}=\cot\alpha\)
c/ \(\tan\alpha.\cot\alpha=\frac{C_đ}{C_k}.\frac{C_k}{C_đ}=1\)
d/ \(\sin^2\alpha=\frac{C_đ^2}{C_h^2}\)
\(\cos^2\alpha=\frac{C_k^2}{C_h^2}\)
\(\Rightarrow\sin^2\alpha+\cos^2\alpha=\frac{C_đ^2+C_k^2}{C_h^2}=\frac{C_h^2}{C_h^2}=1\)
P/s: hok trc lp 9 hay sao mà lm bài bài này?
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CM các hệ thức sau:
a) \(1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\)
b) \(1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\)
c) \(\cot^2\alpha-\cos^2\alpha=\cot^2\alpha.\cos^2\alpha\)
d) \(\frac{1+\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1-\cos\alpha}\)
\(1+tan^2a=1+\frac{sin^2a}{cos^2a}=\frac{cos^2a+sin^2a}{cos^2a}=\frac{1}{cos^2a}\)
\(1+cot^2a=1+\frac{cos^2a}{sin^2a}=\frac{sin^2a+cos^2a}{sin^2a}=\frac{1}{sin^2a}\)
\(cot^2a-cos^2a=\frac{cos^2a}{sin^2a}-cos^2a=cos^2a\left(\frac{1}{sin^2a}-1\right)=cos^2a\left(\frac{1-sin^2a}{sin^2a}\right)\)
\(=cos^2a\left(\frac{cos^2a}{sin^2a}\right)=cos^2a.cot^2a\)
\(\frac{1+cosa}{sina}=\frac{sina\left(1+cosa\right)}{sin^2a}=\frac{sina\left(1+cosa\right)}{1-cos^2a}=\frac{sina\left(1+cosa\right)}{\left(1-cosa\right)\left(1+cosa\right)}=\frac{sina}{1-cosa}\)
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CMR
a)\(\frac{1+\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1-\cos\alpha}\)
b)\(\frac{\tan\alpha+1}{\tan\alpha-1}=\frac{1+\cot\alpha}{1-\cot\alpha}\)
c) \(\tan^2\alpha-\sin^2\alpha=\tan^2\alpha.\sin^2\alpha\)
d)\(\frac{1-4\sin^2\alpha.\cos^2\alpha}{\left(\sin\alpha-\cos\alpha\right)^2}=\left(\sin\alpha+\cos\alpha\right)^2\)
CMR\(\frac{1-2\cos^2\alpha}{1+2\sin\alpha.\cos\alpha}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
\(\frac{1-2\cos^2\alpha}{1+2\sin\alpha.\cos\alpha}=\frac{\left(\sin\alpha-\cos\alpha\right)\left(\sin\alpha+\cos\alpha\right)}{\left(\sin\alpha+\cos\alpha\right)^2}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
các bạn giải rõ ra hộ mk vs
\(\frac{1-\cos^2\alpha}{1+2\sin\alpha.\cos\alpha}=\frac{\sin^2\alpha+\cos^2\alpha-2\cos^2\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha.\cos\alpha}=\frac{\left(\sin\alpha-\cos\alpha\right)\left(\sin\alpha+\cos\alpha\right)}{\left(\sin\alpha+\cos\alpha\right)^2}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
1/ Cho \(\sin\alpha=0,28.\)Tính \(\cos\alpha\), tg\(\alpha\), cotg\(\alpha\)
2/ Cho góc nhọn \(\alpha\). chứng minh rằng: \(1-2\cos^2\alpha=\sin^4\alpha-\cos^4\alpha\)
Câu 1:
\(\cos a=\sqrt{1-0.28^2}=\dfrac{24}{25}\)
\(\tan a=\dfrac{0.28}{0.96}=\dfrac{7}{24}\)
\(\cot a=\dfrac{1}{\tan a}=\dfrac{24}{7}\)
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Dựng góc nhọn α,biết a) sin α frac{2}{3} b) cos α 0,6...
Đọc tiếp
Dựng góc nhọn α,biết a) sin α = \(\frac{2}{3}\) b) cos α = 0,6 c) tg α =\(\frac{3}{4}\) d) cotg =\(\frac{3}{2}\)
Chứng minh các hệ thức sau:
a) \(\frac{1-cos\alpha}{sin\alpha}=\frac{sin\alpha}{1+cos\alpha}\)
b) \(tan^2\alpha-sin^2\alpha=tan^2\alpha.sin^2\alpha\)
c) \(\frac{1-tan\alpha}{1+tan\alpha}=\frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}\)
a) \(\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos a}\)
\(\Leftrightarrow\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=\sin^2\alpha\)
\(\Leftrightarrow1-\cos^2\alpha=\sin^2\alpha\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=1\)( luôn đúng )
\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)
Biết tan α=3. Tính giá trị các biểu thức sau:
a)\(\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
b)\(\frac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-5\cos\alpha}\)
c)\(\frac{1+2\cos^2\alpha}{\sin^2\alpha-\cos^2\alpha}\)
d)\(\frac{\sin^4\alpha+\cos^4\alpha}{1+\sin^2\alpha}\)
\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)
a/ \(\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{3-1}{3+1}\)
b/ \(\frac{2sina+3cosa}{3sina-5cosa}=\frac{3tana+3}{3tana-5}=\frac{3.3+3}{3.3-5}\)
c/ \(\frac{1+2cos^2a}{1-cos^2a-cos^2a}=\frac{1+2cos^2a}{1-2cos^2a}=\frac{1+2.\frac{1}{10}}{1-2.\frac{1}{10}}\)
d/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)
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