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\(=\left(2cos^2a-\left(1-sin^2a\right)\right).\frac{sin^2a}{cos^2a}=\left(2cos^2a-cos^2a\right)\frac{sin^2a}{cos^2a}\)
\(=\frac{cos^2a.sin^2a}{cos^2a}=sin^2a\)
\(1-cot^4a=\left(1+cot^2a\right)\left(1-cot^2a\right)=\frac{1}{sin^2a}\left(2-\left(1+cot^2a\right)\right)\)
\(=\frac{1}{sin^2a}\left(2-\frac{1}{sin^2a}\right)=\frac{2}{sin^2a}-\frac{1}{sin^4a}\)
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)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
a/ Ta có: \(tan\alpha=5\Rightarrow cot\alpha=\frac{1}{5}\) . Đề: \(\frac{sin\alpha}{sin^3\alpha+cos^3\alpha}=\frac{\frac{1}{sin^2\alpha}}{1+\frac{cos^3\alpha}{sin^3\alpha}}=\frac{1+cot^2\alpha}{1+cot^3\alpha}=\frac{1+\left(\frac{1}{5}\right)^2}{1+\left(\frac{1}{5}\right)^3}=\frac{65}{63}\)
b/ Ta có vế trái \(=\frac{sin^2x+cos^2x+cos^2x-sin^2x+\left(sinx+sin3x\right)}{1+2sinx}=\frac{2cos^2x+2.sin2x.cosx}{1+2sinx}=\frac{2cos^2x+4.sinx.cos^2x}{1+2sinx}=\frac{2cos^2x.\left(1+2sinx\right)}{1+2sinx}=2cos^2x\) ( = vế phải)
\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)
\(\frac{1+sin^2a}{1-sin^2a}=\frac{1+sin^2a}{cos^2a}=\frac{1}{cos^2a}+\frac{sin^2a}{cos^2a}=1+tan^2a+tan^2a=1+2tan^2a\)
\(\frac{cosa}{1+sina}+tana=\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina+sin^2a}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sina}{1+cosa}+\frac{1+cosa}{sina}=\frac{sin^2a+cos^2a+2cosa+1}{\left(1+cosa\right)sina}=\frac{2+2cosa}{\left(1+cosa\right)sina}=\frac{2\left(1+cosa\right)}{\left(1+cosa\right)sina}=\frac{2}{sina}\)
a) \(\dfrac{tan\alpha-tan\beta}{cot\beta-cot\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}}{\dfrac{cos\beta}{sin\beta}-\dfrac{cos\alpha}{sin\alpha}}\)
\(=\dfrac{\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}}{\dfrac{cos\beta sin\alpha-cos\alpha sin\beta}{sin\beta sin\alpha}}\)
\(=\dfrac{sin\beta sin\alpha}{cos\beta cos\alpha}=tan\alpha tan\beta\).
b) \(tan100^o+\dfrac{sin530^o}{1+sin640^o}=tan100^o+\dfrac{sin170^o}{1+sin280^o}\)
\(=-cot10^o+\dfrac{sin10^o}{1-sin80^o}\)\(=\dfrac{-cos10^o}{sin10^o}+\dfrac{sin10^o}{1-cos10^o}\)
\(=\dfrac{-cos10^o+cos^210^o+sin^210^o}{sin10^o\left(1-cos10^o\right)}\) \(=\dfrac{1-cos10^o}{sin10^o\left(1-cos10^o\right)}=\dfrac{1}{sin10^o}\) .
\(\left(sin^2a-1-2cos^2a\right)\frac{sin^2a}{cos^2a}=\left(-cos^2a-2cos^2a\right).\frac{sin^2a}{cos^2a}\)
\(=\frac{-3cos^2a.sin^2a}{cos^2a}=-3sin^2a\)
Đề sai hoặc bạn viết sai đề ở \(-2cos^2a\) trên tử số, phải là dấu "+" mới ra kết quả \(sin^2a\)
Sorry, mk nhầm