\(A=\frac{sin5.sin55.sin65}{cos5.cos55.cos65}=\frac{-\frac{1}{2}\left(cos60-cos50\right)sin65}{\frac{1}{2}\left(cos60+cos50\right)cos65}=\frac{-\frac{1}{2}sin65+cos50.sin65}{\frac{1}{2}cos65+cos50.cos65}\)

\(=\frac{-\frac{1}{2}sin65+\frac{1}{2}sin115+\frac{1}{2}sin15}{\frac{1}{2}cos65+\frac{1}{2}cos115+\frac{1}{2}cos15}=\frac{\frac{1}{2}sin15}{\frac{1}{2}cos15}=tan15\)

\(tan30=\frac{1}{\sqrt{3}}=tan2.15=\frac{2tan15}{1-tan^215}\Rightarrow tan^215+2\sqrt{3}tan15-1=0\)

\(\Rightarrow tan15=2-\sqrt{3}\Rightarrow A=2-\sqrt{3}\)

a) \(sin110^ocos130^otan30^ocot320^o\)
Ta có \(sin110^o>0;cos130^o< 0;tan30^o>0;cot320^o< 0\) nên
\(sin110^ocos130^otan30^ocot320^o>0\).
b) \(sin\left(-50^o\right)tan170^ocos\left(-91^o\right)sin530^o\)
\(=-sin50^otan170^o.cos91^osin170^o\)
Do \(sin50^o>0;tan170^o< 0;cos91^o< 0,sin170^o>0\)
nên \(=-sin50^otan170^o.cos91^osin170^o< 0\)
hay \(sin\left(-50^o\right)tan170^ocos\left(-91^o\right)sin530^o< 0\).

a) Ta có :

\(\sin110^0>0;\cos130^0< 0;\tan30^0>0;cot320^0< 0\)

do đó tích của chúng dương.

b) \(\sin\left(-50^0\right)< 0;tan170^0< 0;\cos\left(-91^0\right)< 0;\sin530^0>0\)

do đó tích của chúng âm.

a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{2}\left(\sqrt{5}-2\right)+4\cdot\left(\sqrt{5}+1\right)\)

\(=\dfrac{-3}{2}\sqrt{5}+3+4\sqrt{5}+4\)

\(=\dfrac{5}{2}\sqrt{5}+7\)

b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)

\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)

\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)

\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)

\(=\dfrac{1}{\sin25^0}-1\)

\(=\dfrac{1-\sin25^0}{\sin25^0}\)

ta có cos60=1/2

sin 60=\(\frac{\sqrt{3}}{2}\)

tan 30=\(\frac{\sqrt{3}}{3}\)

ta thay vào biểu thức trên

=> \(\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}+\frac{1}{\frac{\sqrt{3}}{3}}=2\)

\(\frac{cos60^o}{1+sin60^o}+\frac{1}{tan30^o}=\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}+\frac{1}{\frac{\sqrt{3}}{3}}=\frac{1}{2}.\frac{2}{\sqrt{3}+2}+\sqrt{3}=\frac{1}{\sqrt{3}+2}+\sqrt{3}\)

\(=\frac{2-\sqrt{3}}{4-3}+\sqrt{3}=2-\sqrt{3}+\sqrt{3}=2\)

a)
\(A=cos^230^o-sin^230^o=\left(\dfrac{\sqrt{3}}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2}\);
\(B=cos60^o+sin45^o=\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}\).
Vì vậy \(A< B\).
b)
\(C=\dfrac{2tan30^o}{1-tan^230^o}=\dfrac{2\dfrac{\sqrt{3}}{2}}{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}=\sqrt{3}\).
\(D=\left(-tan135^o\right)tan60^o=-\left(-1\right).\sqrt{3}=\sqrt{3}\).
Vậy \(C=D\).

Lời giải:

a)

\(A=\frac{\sin ^2a-\cos ^2a}{\sin a\cos a}=\frac{\sin a}{\cos a}-\frac{\cos a}{\sin a}=\frac{\sin a}{\cos a}-\frac{1}{\frac{\sin a}{\cos a}}=\tan a-\frac{1}{\tan a}\)

\(=\sqrt{3}-\frac{1}{\sqrt{3}}\)

b)

Sử dụng công thức: \(\sin ^2a+\cos ^2a=1; \cos a=\sin (90-a); \tan a=\cot (90-a)\) ta có:

\(B=\cos ^255^0-\cot 58^0+\frac{\tan 52^0}{\cot 38^0}+\cos ^235^0+\tan 32^0\)

\(=\sin ^2(90^0-55^0)-\tan (90^0-58^0)+\frac{\tan 52^0}{\tan (90^0-38^0)}+\cos ^235^0+\tan 32^0\)

\(=(\sin ^235^0+\cos ^235^0)-\tan 32^0+\tan 32^0+\frac{\tan 52^0}{\tan 52^0}\)

\(=1+0+1=2\)

Lời giải:

a)

\(A=\frac{\sin ^2a-\cos ^2a}{\sin a\cos a}=\frac{\sin a}{\cos a}-\frac{\cos a}{\sin a}=\frac{\sin a}{\cos a}-\frac{1}{\frac{\sin a}{\cos a}}=\tan a-\frac{1}{\tan a}\)

\(=\sqrt{3}-\frac{1}{\sqrt{3}}\)

b)

Sử dụng công thức: \(\sin ^2a+\cos ^2a=1; \cos a=\sin (90-a); \tan a=\cot (90-a)\) ta có:

\(B=\cos ^255^0-\cot 58^0+\frac{\tan 52^0}{\cot 38^0}+\cos ^235^0+\tan 32^0\)

\(=\sin ^2(90^0-55^0)-\tan (90^0-58^0)+\frac{\tan 52^0}{\tan (90^0-38^0)}+\cos ^235^0+\tan 32^0\)

\(=(\sin ^235^0+\cos ^235^0)-\tan 32^0+\tan 32^0+\frac{\tan 52^0}{\tan 52^0}\)

\(=1+0+1=2\)

gần off r mới đăng ==" 

sao ko bảo sớm. mấy khi cậu onl.. chắc 1 năm 1 lần. thấy cậu hay lên olm  nên tôi mới bắt đầu lên lại đấy chứ

hơ...tui có sức hút tới v hả, giờ mới biết à, để tối nay chém nha