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\(\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina\left(1+sina\right)}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sin^2a+cos^2a+2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina+cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina+cosa}{sina-cosa}=\frac{\frac{sina}{cosa}+1}{\frac{sina}{cosa}-1}=\frac{tana+1}{tana-1}\)
\(\left(sin^2a\right)^3+\left(cos^2a\right)^3=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)
\(=1-3sin^2a.cos^2a\)
\(sin^2a-tan^2a=tan^4a\left(\frac{sin^2a}{tan^4a}-\frac{1}{tan^2a}\right)=tan^4a\left(sin^2a.\frac{cos^2a}{sin^2a}-\frac{1}{tan^2a}\right)\)
\(=tan^4a\left(cos^2a-cot^2a\right)\) bạn ghi sai đề câu này
\(\frac{tan^3a}{sin^2a}-\frac{1}{sina.cosa}+\frac{cot^3a}{cos^2a}=tan^3a\left(1+cot^2a\right)-\frac{1}{sina.cosa}+cot^3a\left(1+tan^2a\right)\)
\(=tan^3a+tana-\frac{1}{sina.cosa}+cot^3a+cota\)
\(=tan^3a+cot^3a+\frac{sina}{cosa}+\frac{cosa}{sina}-\frac{1}{sina.cosa}\)
\(=tan^3a+cot^3a+\frac{sin^2a+cos^2a-1}{sina.cosa}=tan^3a+cot^3a\)
\(\frac{tan^3a}{sin^2a}-\frac{1}{sina.cosa}+\frac{cot^3a}{cos^2a}=\frac{1}{sin^2a}\left(tan^3a-tana+cot^3a.tan^2a\right)\)
\(=\frac{1}{sin^2a}\left(tan^3a-tana+cota\right)=\left(1+cot^2a\right)\left(tan^3a-tana+cota\right)\)
\(=tan^3a-tana+cota+cot^2a.tan^3a-cot^2a.tana+cot^3a\)
\(=tan^3a-tana+cota+tana-cota+cot^3a\)
\(=tan^3a+cot^3a\)
Lời giải:
a)
\(\frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{(\sin ^2a+\cos ^2a)+\cos ^2a-1}{\cot ^2a}=\frac{1+\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{(\frac{\cos a}{\sin a})^2}=\sin ^2a\)
b)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{\sin ^2a+\cos ^2a}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\tan ^2a+1-1=\tan ^2a\)
c)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}=\frac{\sin ^4a(\cos ^2a-1)}{\cos ^4a(\sin ^2a-1)}\)
\(=\frac{\sin ^4a(-\sin ^2a)}{\cos ^4a(-\cos ^2a)}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)
\(=\frac{1-2sin^2\frac{a}{2}-1}{2cos^2\frac{a}{2}-1+1}=\frac{-sin^2\frac{a}{2}}{cos^2\frac{a}{2}}=-tan^2\frac{a}{2}\)
\(\frac{sin^4x-sin^2x+cos^2x}{cos^4x-cos^2x+sin^2x}=\frac{sin^2x\left(sin^2x-1\right)+cos^2x}{cos^2x\left(cos^2x-1\right)+sin^2x}=\frac{-sin^2x.cos^2x+cos^2x}{-cos^2x.sin^2x+sin^2x}\)
\(=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)
\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)
\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)
\(A=\frac{sin^2a-tan^2a}{cos^2a-cot^2a}=\frac{sin^2a-\frac{sin^2a}{cos^2a}}{cos^2a-\frac{cos^2a}{sin^2a}}=\frac{\frac{sin^2a\left(cos^2a-1\right)}{cos^2a}}{\frac{cos^2a\left(sin^2a-1\right)}{sin^2a}}=\frac{sin^4a.\left(-sin^2a\right)}{cos^4a.\left(-cos^2a\right)}=\frac{sin^6a}{cos^6a}=tan^6a\)
\(\dfrac{\sin^2a-\tan^2a}{\cos^2a-\cot^2a}=\dfrac{\sin^2a-\dfrac{\sin^2a}{\cos^2a}}{\cos^2a-\dfrac{\cos^2a}{\sin^2a}}=\dfrac{\dfrac{\sin^2a\cos^2a-\sin^2a}{\cos^2a}}{\dfrac{\cos^2a\sin^2a-\cos^2a}{\sin^2a}}=\dfrac{\sin^2a\sin^2a\left(\cos^2a-1\right)}{\cos^2a\cos^2a\left(\sin^2a-1\right)}\)
\(=\dfrac{\sin^4a\left(\cos^2a-\cos^2a-\sin^2a\right)}{\cos^4a\left(\sin^2a-\cos^2a-\sin^2a\right)}=\dfrac{\sin^4a\left(-\sin^2a\right)}{\cos^4a\left(-\cos^2a\right)}\)
\(=\dfrac{-\sin^6a}{-\cos^6a}=\dfrac{\sin^6a}{\cos^6a}=\tan^6a\)
Lời giải:
a)
\(\frac{\cos (a-b)}{\cos (a+b)}=\frac{\cos a\cos b+\sin a\sin b}{\cos a\cos b-\sin a\sin b}=\frac{\frac{\cos a\cos b}{\sin a\sin b}+1}{\frac{\cos a\cos b}{\sin a\sin b}-1}=\frac{\cot a\cot b+1}{\cot a\cot b-1}\)
b)
\(2(\sin ^6a+\cos ^6a)+1=2(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+1\)
\(=2(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+1\)
\(=3(\sin ^4a+\cos ^4a)-(\sin ^4a+\cos ^4a+2\sin ^2a\cos ^2a)+1\)
\(=3(\sin ^4a+\cos ^4a)-(\sin ^2a+\cos ^2a)^2+1\)
\(=3(\sin ^4a+\cos ^4a)-1^2+1=3(\sin ^4a+\cos ^4a)\)
c)
\(\frac{\tan a-\tan b}{cot b-\cot a}=\frac{\tan a-\tan b}{\frac{1}{\tan b}-\frac{1}{\tan a}}\) (nhớ rằng \(\tan x.\cot x=1\rightarrow \cot x=\frac{1}{\tan x}\) )
\(=\frac{\tan a-\tan b}{\frac{\tan a-\tan b}{\tan a\tan b}}=\tan a\tan b\)
d)
\((\cot x+\tan x)^2-(\cot x-\tan x)^2=(\cot ^2x+\tan ^2x+2\cot x\tan x)-(\cot ^2x-2\cot x\tan x+\tan ^2x)\)
\(=4\cot x\tan x=4.1=4\)
e)
\(\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)
\(=\sin ^2a-\sin a\cos a+\cos ^2a=(\sin ^2a+\cos ^2a)-\sin a\cos a=1-\sin a\cos a\)
Vậy ta có đpcm.
\(D=\frac{\frac{sina}{cos^3a}+\frac{5cosa}{cos^3a}}{\frac{sin^3a}{cos^3a}-\frac{2cos^3a}{cos^3a}}=\frac{tana.\frac{1}{cos^2a}+\frac{5}{cos^2a}}{tan^3a-2}=\frac{tana\left(1+tan^2a\right)+5\left(1+tan^2a\right)}{tan^3a-2}\)
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Lời giải:
Ta có:
\(\frac{\tan ^3a}{\sin ^2a}-\frac{1}{\sin a\cos a}+\frac{\cot ^3a}{\cos ^2a}=\frac{\tan ^3a\cos ^2a+\cot ^3a\sin ^2a}{\sin ^2a\cos ^2a}-\frac{\sin a\cos a}{\sin ^2a\cos ^2a}\)
\(=\frac{\frac{\sin ^3a}{\cos ^3a}.\cos ^2a+\frac{\cos ^3a}{\sin ^3a}.\sin ^2a}{\sin ^2a\cos ^2a}-\frac{\sin a\cos a}{\sin ^2a\cos ^2a}\)
\(=\frac{\frac{\sin ^3a}{\cos a}+\frac{\cos ^3a}{\sin a}-\sin a\cos a}{\sin ^2a\cos ^2a}=\frac{\sin ^4a+\cos ^4a-\sin ^2a\cos ^2a}{\sin ^3a\cos ^3a}\)
\(=\frac{(\sin ^2a+\cos ^2a)(\sin ^4a+\cos ^4a-\sin ^2a\cos ^2a)}{\sin ^3a\cos ^3a}\)
\(=\frac{\sin ^6a+\cos ^6a}{\sin ^3a\cos ^3a}=\frac{\sin ^3a}{\cos ^3a}+\frac{\cos ^3a}{\sin ^3a}=\tan ^3a+\cot ^3a\)
Ta có đpcm.