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

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)

\(\RightarrowĐPCM\)

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

\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)

a, \(\left(1-sin^2x\right)cot^2x+1-cot^2x\)

\(=cot^2x-sin^2x.cot^2x+1-cot^2x\)

\(=1-sin^2x.\frac{\text{cos}^2x}{sin^2x}=1-\text{cos}^2x=sin^2x\)

b,\(\left(tanx+cotx\right)^2-\left(tanx-cotx\right)2\)

\(=tan^2x2.tanx.cotx+cot^2x-tan^2x+2tanx.cotx-cot^2x\)

\(=4tanxcotx=4\)

c,\(\left(xsina-y\text{cos}a\right)^2+\left(x\text{cos}a+ysina\right)^2\)

\(=x^2sin^2a=2xysina\text{cos}a+y^2\text{cos}^2a+2xysina\text{cos}a+y^2sin^2a\)

\(=x^2\left(sin^2a+\text{cos}^2a\right)+y^2\left(sin^2a+\text{cos}^2a\right)\)

\(=x^2+y^2\)

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.

\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)

\(=tan^2x+1+tan^2x=1+2tan^2x\)

\(a)sin^4x+cos^4x=1-2sin^2x\cdot cos^2x\) 

\(\Leftrightarrow sin^4x+2sin^2x\cdot cos^2x+cos^4x=1\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2=1\)(luôn đúng)

a) \sin ^{4} x+\cos ^{4} x=\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} x-2 \sin ^{2} x \cos ^{2} xsin4x+cos4x=sin4x+cos4x+2sin2xcos2x−2sin2xcos2x
\begin{aligned}&=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\&=1-2 \sin ^{2} x \cos ^{2} x\end{aligned}​=(sin2x+cos2x)2−2sin2xcos2x=1−2sin2xcos2x​

b) \dfrac{1+\cot x}{1-\cot x}=\dfrac{1+\dfrac{1}{\tan x}}{1-\dfrac{1}{\tan x}}=\dfrac{\dfrac{\tan x+1}{\tan x}}{\dfrac{\tan x-1}{\tan x}}=\dfrac{\tan x+1}{\tan x-1}1−cotx1+cotx​=1−tanx1​1+tanx1​​=tanxtanx−1​tanxtanx+1​​=tanx−1tanx+1​

c) \dfrac{\cos x+\sin x}{\cos ^{3} x}=\dfrac{1}{\cos ^{2} x}+\dfrac{\sin x}{\cos ^{3} x}=\tan ^{2} x+1+\tan x\left(\tan ^{2} x+1\right)cos3xcosx+sinx​=cos2x1​+cos3xsinx​=tan2x+1+tanx(tan2x+1)
=\tan ^{3} x+\tan ^{2} x+\tan x+1=tan3x+tan2x+tanx+1

a) VT=(sin²x + cos ² x)² - 2sin² x . cos² x = VP 

b) VT= \(\dfrac{1+\dfrac{1}{tanx}}{1-\dfrac{1}{tanx}}\)=VP

c) VT= \(\dfrac{1}{cos^2x}+\dfrac{sinx}{cosx}.\dfrac{1}{cos^2x}=1+tan^2x+tanx.\left(1+tan^2x\right)=VP\)

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 ^^

a)theo tỉ số lượng giác ta có: tan a= AC/AB (1)

sin a= AC/BC

cos a= AB/BC

-> sin a * cos a= AC/BC : BC/AB= AC/AB (2)

Từ (1) (2) ta có tan a = sina / cos a

bạn có cần gấp ko

làm tương tự nhé và nhớ tích cho mk nhé