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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: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)
\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)
\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)
\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)
\(=\dfrac{1}{\sin25^0}-1\)
\(=\dfrac{1-\sin25^0}{\sin25^0}\)
tan(2x+10o)+cot(x)=0
<=> tan(2x+10o)+tan(90o-x)=0
<=>tan(x+100o)*[1-tan(2x-10o)*tan(90o-x)]=0
*tan(x+100o)=0 => x=....
*1-tan(2x-10o)*tan(90o-x)=0
<=> tan(2x-10o)=tanx <=> x=....
Bổ sung đề: \(0^0 . Tính sin x, cosx
tan x-cot x=3/2
=>\(\tan x-\frac{1}{tanx}=\frac32\)
=>\(\frac{\tan^2x-1}{\tan x}=\frac32\)
=>\(2\cdot\tan^2x-2=3\cdot\tan x\)
=>\(2\cdot\tan^2x-3\cdot\tan x-2=0\)
=>\(\left(2\cdot\tan^2x-4\cdot\tan x\right)+\left(\tan x-2\right)=0\)
=>(tan x-2)(2*tan x+1)=0
=>tan x-2=0(Vì tan x>0 do \(0^0 )
=>tan x=2
\(1+\tan^2x=\frac{1}{cos^2x}\)
=>\(\frac{1}{cos^2x}=1+2^2=5\)
=>\(cos^2x=\frac15\)
=>\(cosx=\frac{1}{\sqrt5}\)
\(\sin x=cosx\cdot\tan x\)
\(=2\cdot\frac{1}{\sqrt5}=\frac{2}{\sqrt5}\)
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 \(C=D\).
a) Ta có: \(sin^2x+sin^2\left(90-x\right)=sin^2x+cos^2x=1.\)
áp dụng: A = 2
b)Ta có: \(cos\left(x\right)=-cos\left(180-x\right)\)
áp dụng: B = 0
c) Ta có: \(tan\left(x\right)\cdot tan\left(90-x\right)=\frac{sinx}{cosx}\cdot\frac{sin\left(90-x\right)}{cos\left(90-x\right)}=\frac{sinx}{cosx}\cdot\frac{cosx}{sinx}=1\)
áp dụng: C = 1
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) \(tan100^o+\dfrac{sin530^o}{1+sin640^o}=tan100^o+\dfrac{sin170^o}{1+sin280^o}\)
\(=-cot10^o+\dfrac{sin10^o}{1-sin80^o}\)\(=\dfrac{-cos10^o}{sin10^o}+\dfrac{sin10^o}{1-cos10^o}\)
\(=\dfrac{-cos10^o+cos^210^o+sin^210^o}{sin10^o\left(1-cos10^o\right)}\) \(=\dfrac{1-cos10^o}{sin10^o\left(1-cos10^o\right)}=\dfrac{1}{sin10^o}\) .
Ta có:
tan 89 0 = c o t 1 0 ; tan 88 0 = c o t 2 0 ; …; tan 46 0 = c o t 44 0 và tan α . c o t α = 1
Nên B = tan 1 0 . tan 89 0 . tan 2 0 . tan 88 0 … tan 46 0 . tan 44 0 . tan 45 0
= tan 1 0 . c o t 1 0 . tan 2 0 . c o t 2 0 . tan 3 0 . c o t 3 0 … tan 44 0 . c o t 44 0 . tan 45 0
= 1.1.1….1.1 = 1
Vậy B = 1
Đáp án cần chọn là: B