chứng minh các tslg sau
a) tan α = \(\dfrac{sin a}{cos a}\)
b)cot a = \(\dfrac{cos a}{sin a}\)
c)tan a . cot a = 1
Chứng minh đẳng thức:
a, \(\dfrac{\sin x+\cos x-1}{1-\cos x}=\dfrac{2\cos x}{\sin x-\cos x+1}\)
b, \(\tan a.\tan b=\dfrac{\tan a+\tan b}{\cot a+\cot b}\)
a/ \(\dfrac{\sin x+\cos x-1}{1-\cos x}=\dfrac{2\cos x}{\sin x-\cos x+1}\)
\(\Leftrightarrow-2\cos^2x+2\cos x-2\cos x+2\cos^2x=0\)
\(\Leftrightarrow0=0\) (đúng)
\(\RightarrowĐPCM\)
b/ \(\tan a.\tan b=\dfrac{\tan a+\tan b}{\cot a+\cot b}\)
\(\Leftrightarrow\tan a.\tan b.\left(\cot a+\cot b\right)=\tan a+\tan b\)
\(\Leftrightarrow\tan a.\tan b.\cot a+\tan a.\tan b.\cot b=\tan a+\tan b\)
\(\Leftrightarrow\tan b+\tan a=\tan a+\tan b\) (đúng)
\(\RightarrowĐPCM\)
Giả sử A, B, C là ba góc của tam giác ABC, chứng minh rằng :
a) \(\dfrac{\sin C}{\cos A\cos B}=\tan A+\tan B\)
b) \(\sin A+\sin B+\sin C=4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}\)
c) \(\dfrac{\sin A+\sin B+\sin C}{\sin A+\sin B-\sin C}=\cot\dfrac{A}{2}\cot\dfrac{B}{2}\)
Chứng minh các đẳng thức sau(giả sử các biểu thức sau đều có nghĩa)
a) $\sin ^{4} x+\cos ^{4} x=1-2 \sin ^{2} x \cdot \cos ^{2} x$.
b) $\dfrac{1+\cot x}{1-\cot x}=\dfrac{\tan x+1}{\tan x-1}$.
c) $\dfrac{\cos x+\sin x}{\cos ^{3} x}=\tan ^{3} x+\tan ^{2} x+\tan x+1$.
\(a)sin^4x+cos^4x=1-2sin^2x\cdot cos^2x\)
\(\Leftrightarrow sin^4x+2sin^2x\cdot cos^2x+cos^4x=1\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2=1\)(luôn đúng)
a)
b)
c)
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
\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)
Cho
1.tan α=\(\dfrac{1}{3} \) tính A=\(\dfrac{2\sin^2x+5}{4\cos^2x-3}\)
2.cot α=\(\dfrac{2}{5}\) tính B=\(\dfrac{3\cos^2x-\sin^2x}{c\text{os}^2x+2\sin^2x}\)
\(A=\dfrac{2tan^2a+\dfrac{5}{cos^2a}}{4-\dfrac{3}{cos^2a}}=\dfrac{2tan^2a+5\left(1+tan^2a\right)}{4-3\left(1+tan^2a\right)}=...\) (bạn tự thay số bấm máy nhé)
\(B=\dfrac{3cot^2a-1}{cot^2a+2}=...\)
chứng minh rằng
a) tanx(cot\(^2\)x - 1) = cotx(1 - tan\(^2\)x)
b) tan\(^2\)x - sin\(^2\)x = tan\(^2\)x.sin\(^2\)x
c) \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}\) - cos\(^2\)x = - cos\(^4\)x
a: tan x(cot^2x-1)
\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)
=cotx-tanx/cotx=cotx(1-tan^2x)
b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)
\(=sin^2x\left(\dfrac{1}{cos^2x}-1\right)=sin^2x\cdot\dfrac{sin^2x}{cos^2x}=sin^2x\cdot tan^2x\)
c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)
\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x\cdot cos^2x}{1}=sin^2x\cdot cos^2x\)
=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)
=-cos^2x*cos^2x=-cos^4x
=>ĐPCM
Chứng minh các đẳng thức sau:
a) \(\dfrac{1}{1+\tan\alpha}+\dfrac{1}{1+\cot\alpha}=1\) b) \(\sin^4x-\cos^4x=2\sin^2x-1\)
c) \(\dfrac{1}{\sin^2x}+\dfrac{1}{\cos^2x}=\tan^2x+\cot^2x+2\)
d) \(\sin x.\cos x.\left(1+\tan x\right)\left(1+\cot x\right)=1+2\sin x\)
a) \(\dfrac{1}{1+tan\alpha}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{1}{1+\dfrac{1}{cot\alpha}}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{1}{\dfrac{cot\alpha+1}{cot\alpha}}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{cot\alpha}{cot\alpha+1}+\dfrac{1}{1+cot\alpha}\)
\(=\dfrac{cot\alpha+1}{cot\alpha+1}=1\) (đpcm)
b) \(tan^2x+cot^2x+2\)
\(=\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{sin^2x}+2\)
\(=\dfrac{sin^2x}{cos^2x}+1+\dfrac{cos^2x}{sin^2x}+1\)
\(=\dfrac{sin^2x+cos^2x}{cos^2x}+\dfrac{cos^2x+sin^2x}{sin^2x}\)
\(=\dfrac{1}{cos^2x}+\dfrac{1}{sin^2x}\) (đpcm)
c) \(sinx.cosx.\left(1+tanx\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sinx.cosx.tanx\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sinx.cosx.\dfrac{sinx}{cosx}\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sin^2x\right)\left(1+cotx\right)\)
\(=\left(sinx.cosx+sin^2x\right)\left(1+\dfrac{cosx}{sinx}\right)\)
\(=sinx.cosx+cos^2x+sin^2x+sinx.cosx\)
\(=1+sin^2x.cos^2x\)
Câu cuối không biết chỗ sai, mong mọi người chỉ bảo ạ ^^
Chứng minh:
1.\(\dfrac{\cot^2x-\sin^2x}{\cot^2x-\tan^2x}=\sin^2x\cdot\cos^2x\)
2.\(\dfrac{1-\sin x}{\cos x}-\dfrac{\cos x}{1+\sin x}=0\)
3.\(\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cot x}=\cos x\)
4.\(\dfrac{\tan x}{1-\tan^2x}\cdot\dfrac{\cot^2x-1}{\cot x}=1\)
5.\(\dfrac{1+\sin^2x}{1-\sin^2x}=1+2\tan^2x\)
Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)
2.
\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)
\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)
3.
\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)
4.
\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)
5.
\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)
\(=tan^2x+1+tan^2x=1+2tan^2x\)
Câu 50**: Cho góc nhọn α tuỳ ý giá trị biểu thức \(\dfrac{tan\alpha}{cot\alpha}+\dfrac{cot\alpha}{tan\alpha}-\dfrac{sin^2\alpha}{cos^2\alpha}\)bằng
A. \(tan^2\alpha\) ; B . \(cot^2\) α ; C . 0 ; D. 1 .
giải hộ mik vs