Rút gọn biểu thức:
\(B=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)-\left(1+cot^2\alpha\right)\left(1-cos^2\alpha\right)\)
Rút gọn các biểu thức sau:
a) \(\frac{1}{{\tan \alpha + 1}} + \frac{1}{{\cot \alpha + 1}}\)
b) \(\cos \left( {\frac{\pi }{2} - \alpha } \right) - \sin \left( {\pi + \alpha } \right)\)
c) \(\sin \left( {\alpha - \frac{\pi }{2}} \right) + \cos \left( { - \alpha + 6\pi } \right) - \tan \left( {\alpha + \pi } \right)\cot \left( {3\pi - \alpha } \right)\)
\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)
\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)
\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)
Rút gọn các biểu thức:
a)\(\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
b)\(\cot^2\alpha-\cos^2\alpha.\cot^2\alpha\)
c)\(\sin\alpha.\cos\alpha\left(\tan\alpha+\cot\alpha\right)\)
d)\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha\)
a) khai triển được 2sin2+2cos2=2(sin2+cos2=2.1=2
b)cot2-cos2.cot2=cot2(1-cos2)=cot2.sin2=cos2/sin2.sin2=cos2
c)sin.cos(tan+cot)=sin.cos.tan+sin.cos.cot=sin.cos.sin/cos+sin.cos.cos/sin=sin2+cos2=1
d)tan2-tan2.sin2=tan2(1-sin2)=tan2.cos2=sin2/cos2.cos2=sin2
Rút gọn biểu thức:
\(A=\sin^210+\sin^220+\sin^230+\sin^280+\sin^270+\sin^260\)
\(B=\left(1+\tan^2\alpha\right)\left(1-\sin^2\alpha\right)+\left(1+\cot^2\alpha\right)\left(1-\cos^2\alpha\right)\)
\(A=sin^210+sin^220+sin^230+sin^280+sin^270+sin^260=sin^210+sin^220+sin^230+cos^210+cos^220+cos^230=1+1+1=3\)\(B=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)+\left(1+cot^2\alpha\right)\left(1-cos^2\alpha\right)=\dfrac{1}{cos^2\alpha}.cos^2\alpha+\dfrac{1}{sin^2\alpha}.sin^2\alpha=1+1=2\)
đơn giản biểu thức:
a, \(\left(\frac{sin\alpha+tan\alpha}{cos\alpha+1}\right)^2+1\)
b, \(tan\alpha\left(\frac{1+cos^2\alpha}{sin\alpha}-sin\alpha\right)\)
c, \(\frac{cot^2\alpha-cos^2\alpha}{cot^2a}+\frac{sin\alpha.cos\alpha}{cot\alpha}\)
\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)
1. Đơn giản biểu thức
a. \(\sin\alpha\cdot\cos\alpha\left(\tan\alpha+\cot\alpha\right)\)
b. \(\left(\sin^2\alpha+\cos^2\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
c. \(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\)
a) ta có : \(sin\alpha.cos\alpha\left(tan\alpha+cot\alpha\right)=sin\alpha.cos\alpha\left(\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}\right)\)
\(=sin^2\alpha+cos^2\alpha=1\)
b) ta có : \(\left(sin^2\alpha+cos^2\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1^2+1-2sin\alpha.cos=2\left(1-2sin\alpha.cos\alpha\right)\)
c) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
rút gọn
a)A=\(\frac{1+2cos\alpha.sin\alpha}{cos^2\alpha-sin^2\alpha}\)
b)B=\(\left(1+\cot^2\alpha\right)\left(1-sin^2\alpha\right)\)-\(\left(1+\cot^2\alpha\right)\left(1-\cos^2\alpha\right)\)
c)C=\(\sin^6\alpha+\cos^6\alpha\)+\(3\sin^2\alpha.cos^2\alpha\)
rút gọn:
\(\left(1+\tan^2\alpha\right)\left(1-\sin^2\alpha\right)-\left(1+\cot^2\alpha\right)\left(1-\cos^2\alpha\right)\)
Lời giải:
Gọi biểu thức trên là $P$. Ta có:
$P=(1+\frac{\sin ^2a}{\cos ^2a})(1-\sin ^2a)-(1+\frac{\cos ^2a}{\sin ^2a})(1-\cos ^2a)$
$=\frac{\sin ^2a+\cos ^2a}{\cos ^2a}.\cos ^2a-\frac{\sin ^2a+\cos ^2a}{\sin ^2a}.\sin ^2a$
$=(\sin ^2a+\cos ^2a)-(\sin ^2a+\cos ^2a)$
$=0$
H= \(\sin^4\alpha\left(1+2\cos^2\alpha\right)+\cos^4\alpha\left(1+2\sin^2\alpha\right)\)
K=\(\tan^2\alpha\left(2\cos^2\alpha+\sin^2\alpha-1\right)\)
Rút gọn biểu thức
Đơn giản các biểu thức sau:
a) \(\sin {100^o} + \sin {80^o} + \cos {16^o} + \cos {164^o};\)
b) \(2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) với \({0^o} < \alpha < {90^o}\).
a) Ta có: \(\left\{ \begin{array}{l}\sin {100^o} = \sin \left( {{{180}^o} - {{80}^o}} \right) = \sin {80^o}\\\cos {164^o} = \cos \left( {{{180}^o} - {{16}^o}} \right) = - \cos {16^o}\end{array} \right.\)
\( \Rightarrow \sin {100^o} + \sin {80^o} + \cos {16^o} + \cos {164^o}\)\( = \sin {80^o} + \sin {80^o} + \cos {16^o}-\cos {16^o}\)\( = 2\sin {80^o}.\)
b)
Ta có:
\(\left\{ \begin{array}{l}\sin \left( {{{180}^o} - \alpha } \right) = \sin \alpha \\\cos \left( {{{180}^o} - \alpha } \right) = - \cos \alpha \\\tan \left( {{{180}^o} - \alpha } \right) = - \tan \alpha \\\cot \left( {{{180}^o} - \alpha } \right) = - \cot \alpha \end{array} \right.\quad ({0^o} < \alpha < {90^o})\)\( \Rightarrow 2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) \( = 2\sin \alpha .\cot \alpha - \left( { - \cos \alpha } \right).\tan \alpha .\left( { - \cot \alpha } \right)\)\( = 2\sin \alpha .\cot \alpha - \cos \alpha .\tan \alpha .\cot \alpha \)
\( = 2\sin \alpha .\frac{{\cos \alpha }}{{\sin \alpha }} - \cos \alpha .\left( {\tan \alpha .\cot \alpha } \right)\)\( = 2\cos \alpha - \cos \alpha .1 = \cos \alpha .\)