CMR: Sin a < tan a;cos a < cot a
cmr: tan a/2 tan b/2 tan c/2 lập thành cấp số nhân khi cos a/2 cos b/2 cos c/2 là cấp số nhân.
cmr: cot a/2 cot b/2 cot c/2 là csn khi sin a/2 sin b/2 sin c/2 là cấp số nhân
CMR sin a < tan a
cos a < cot a
1.cho góc nhọn a cmr: sin a<tan a;cos a<cot a
2.cmr:1+\(tan^2\)a=\(\frac{1}{cos^2a}\);1+\(cot^2\)a=\(\frac{1}{sin^2a}\)
1.
$0< a< 90^0\Rightarrow `1>\sin a, \cos a>0$
Do đó:
$\sin a-\tan a=\sin a-\frac{\sin a}{\cos a}=\frac{\sin a(\cos a-1)}{\cos a}<0$
$\Rightarrow \sin a< \tan a$
(đpcm)
$\cos a-\cot a=\cos a-\frac{\cos a}{\sin a}=\frac{\cos a(\sin a-1)}{\sin a}<0$
$\Rightarrow \cos a< \cot a$ (đpcm)
Bài 2:
\(1+\tan ^2a=1+\frac{\sin ^2a}{\cos ^2a}=\frac{\cos ^2a+\sin ^2a}{\cos ^2a}=\frac{1}{\cos ^2a}\)
\(1+\cot ^2a=1+\frac{\cos ^2a}{\sin ^2a}=\frac{\sin ^2a+\cos ^2a}{\sin ^2a}=\frac{1}{\sin ^2a}\)
Ta có đpcm.
1. a) CMR : A =\(\frac{1-2\sin\alpha.\cos\alpha}{sin^2\alpha-cos^2\alpha}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
b) Tính A khi \(\tan\alpha\) =\(\frac{1}{3}\)
CMR
a)\(\frac{1+\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1-\cos\alpha}\)
b)\(\frac{\tan\alpha+1}{\tan\alpha-1}=\frac{1+\cot\alpha}{1-\cot\alpha}\)
c) \(\tan^2\alpha-\sin^2\alpha=\tan^2\alpha.\sin^2\alpha\)
d)\(\frac{1-4\sin^2\alpha.\cos^2\alpha}{\left(\sin\alpha-\cos\alpha\right)^2}=\left(\sin\alpha+\cos\alpha\right)^2\)
CMR:
a, \(\frac{\cot^2x-\sin^2x}{\cot^2x-tan^2x}=sin^2x.\cos^2x\)
b, \(\frac{\tan x}{1-\tan^2x}.\frac{\cot^2-1}{\cot x}=1\)
c, \(\frac{1+\sin x.\cos x}{\sin^2x-\cos^2x}=\frac{\tan x+1}{\cot x+1}\)
d, \(\frac{\sin x+\cos x-1}{\sin x-cosx+1}=\frac{\cos x}{1+sinx}\)
cho tg ABC\(\perp\)A, đường phân giác BD.
CMR: a) \(\tan\dfrac{B}{2}=\dfrac{AC}{BC+AB}\)
CMR: b) S(ABC)=\(\dfrac{AB\times BC}{2}\times\sin B\)
b: \(\dfrac{AB\cdot BC}{2}\cdot sinB\)
\(=\dfrac{AB\cdot BC}{2}\cdot\dfrac{AC}{BC}=\dfrac{AB\cdot AC}{2}\)
\(=S_{ABC}\)
a: Xét ΔABD vuông tại A có tan ABD=AD/AB
Xét ΔCBA có BD là phân giác
nên AD/AB=CD/BC
=>\(\dfrac{AD}{AB}=\dfrac{CD}{BC}=\dfrac{AD+CD}{AB+BC}=\dfrac{AC}{AB+BC}\)
=>\(tan\left(ABD\right)=\dfrac{AC}{AB+BC}\)
Cho tam giác ABC vuông tại A. CMR
a, \(\sin B< 1;\cos B< 1\)
b, \(\tan B=\frac{\sin B}{\cos B}\)
c, \(\cot B=\frac{\cos B}{\sin B}\)
d, \(\tan B.\cot B=1\)
e, \(\sin^2B+\cos^2B=1\)
\(\cot B=\frac{\cos B}{\sin B}\)
a) Biết sin a =\(\dfrac{2}{3}\).Tính cos a,tan a,cot a
b)Biết cos a =\(\dfrac{1}{5}\).Tính sin a, tan a,cot a
c)Biết tan a = 2.Tính sin a,cos a ,cot a.
a: sin a=2/3
=>cos^2a=1-(2/3)^2=5/9
=>\(cosa=\dfrac{\sqrt{5}}{3}\)
\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)
\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
b: cos a=1/5
=>sin^2a=1-(1/5)^2=24/25
=>\(sina=\dfrac{2\sqrt{6}}{5}\)
\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)
\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)
c: cot a=1/tana=1/2
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>1/cos^2a=1+4=5
=>cos^2a=1/5
=>cosa=1/căn 5
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)