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\(1+tan^2a=1+\frac{sin^2a}{cos^2a}=\frac{cos^2a+sin^2a}{cos^2a}=\frac{1}{cos^2a}\)
\(1+cot^2a=1+\frac{cos^2a}{sin^2a}=\frac{sin^2a+cos^2a}{sin^2a}=\frac{1}{sin^2a}\)
\(cot^2a-cos^2a=\frac{cos^2a}{sin^2a}-cos^2a=cos^2a\left(\frac{1}{sin^2a}-1\right)=cos^2a\left(\frac{1-sin^2a}{sin^2a}\right)\)
\(=cos^2a.\frac{cos^2a}{sin^2a}=cos^2a.cot^2a\)
Câu cuối đề bài sai
\(1+tan^2a=1+\frac{sin^2a}{cos^2a}=\frac{cos^2a+sin^2a}{cos^2a}=\frac{1}{cos^2a}\)
\(1+cot^2a=1+\frac{cos^2a}{sin^2a}=\frac{sin^2a+cos^2a}{sin^2a}=\frac{1}{sin^2a}\)
\(cot^2a-cos^2a=\frac{cos^2a}{sin^2a}-cos^2a=cos^2a\left(\frac{1}{sin^2a}-1\right)=cos^2a\left(\frac{1-sin^2a}{sin^2a}\right)\)
\(=cos^2a\left(\frac{cos^2a}{sin^2a}\right)=cos^2a.cot^2a\)
\(\frac{1+cosa}{sina}=\frac{sina\left(1+cosa\right)}{sin^2a}=\frac{sina\left(1+cosa\right)}{1-cos^2a}=\frac{sina\left(1+cosa\right)}{\left(1-cosa\right)\left(1+cosa\right)}=\frac{sina}{1-cosa}\)
a) \(\frac{1+2sina.cosa}{cos^2a-sin^2a}=\frac{1+sin2a}{cos2a}\)
b) \(B=\left(1+tan^2a\right)\left(1-sin^2a\right)-\left(1+cot^2a\right)\left(1-cos^2a\right)\)
\(=\left(1+\frac{sin^2a}{cos^2a}\right)\left(sin^2a+cos^2a-sin^2a\right)-\left(1+\frac{cos^2a}{sin^2a}\right)\left(cos^2a+sin^2a-cos^2a\right)\)
\(=\left(\frac{cos^2a+sin^2a}{cos^2a}\right).cos^2a-\left(\frac{sin^2a+cos^2a}{sin^2a}\right).sin^2a\)
\(=\frac{1}{cos^2a}.cos^2a-\frac{1}{sin^2a}.sin^2a=1-1=0\)
c)
\(C=\left(sin^2a+cos^2a\right)^3-3.sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)
\(=1-3sin^2a.cos^2a\left(1-1\right)=1\)
+) ta có : \(A=\left(tan\alpha+cot\alpha\right)^2-\left(tan\alpha-cot\alpha\right)^2\)
\(=tan^2\alpha+cot^2\alpha+2-tan^2\alpha-cot^2\alpha+2=4\) (không phụ thuộc vào \(\alpha\)) \(\Rightarrow\) (đpcm)
+) ta có : \(B=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
\(=\left(sin^2\alpha\right)^3+\left(cos^2\alpha\right)^3+3sin^2\alpha.cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha.cos^2\alpha\)
\(=\left(\left(sin^2\alpha+cos^2\alpha\right)^2-3sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha.cos^2\alpha\)
\(=1\) (không phụ thuộc vào \(\alpha\) ) \(\Rightarrow\) (đpcm)
Lời giải:
a) Áp dụng công thức \(\sin ^2a+\cos ^2a=1\) thì:
\(P=3\sin ^2a+4\cos ^2a=3(\sin ^2a+\cos ^2a)+\cos ^2a\)
\(=3.1+(\frac{1}{3})^2=\frac{28}{9}\)
b)
\(\tan a=\frac{3}{4}\Rightarrow \cot a=\frac{1}{\tan a}=\frac{4}{3}\)
\(\frac{3}{4}=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\frac{3}{4}\cos a\)
\(\Rightarrow \sin ^2a=\frac{9}{16}\cos ^2a\)
\(\Rightarrow \sin ^2a+\cos ^2a=\frac{25}{16}\cos ^2a\Rightarrow \frac{25}{16}\cos ^2a=1\)
\(\Rightarrow \cos ^2a=\frac{16}{25}\Rightarrow \cos a=\pm \frac{4}{5}\)
Nếu \(\Rightarrow \sin a=\pm \frac{3}{5}\) (theo thứ tự)
c)
\(\frac{1}{2}=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\frac{\cos a}{2}\). Vì a góc nhọn nên \(\cos a\neq 0\)
Do đó:
\(\frac{\cos a-\sin a}{\cos a+\sin a}=\frac{\cos a-\frac{\cos a}{2}}{\cos a+\frac{\cos a}{2}}=\frac{\cos a(1-\frac{1}{2})}{\cos a(1+\frac{1}{2})}=\frac{1-\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}\)
a) \(1+tan^2\alpha=1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2=\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha}\)
b) \(1+cot^2\alpha=1+\left(\dfrac{cos\alpha}{sin\alpha}\right)^2=\dfrac{cos^2\alpha+sin^2\alpha}{sin^2\alpha}=\dfrac{1}{sin^2\alpha}\)
c) \(tan^2\alpha\left(2sin^2\alpha+3cos^2\alpha-2\right)=tan^2\alpha\left[cos^2\alpha+2\left(sin^2\alpha+cos^2\alpha\right)-2\right]=\dfrac{sin^2\alpha}{cos^2\alpha}\times cos^2\alpha=sin^2\alpha\)
a)
\(1+tan^2\alpha=1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2=\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha}\)
b)\(1+cot^2\alpha=1+\left(\dfrac{cos\alpha}{sin\alpha}\right)^2=\dfrac{sin^2\alpha+cos^2\alpha}{sin^2\alpha}=\dfrac{1}{sin^2\alpha}\)
c) mình chưa rõ đề nha