Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ta có : \(sin\alpha.cos\alpha\left(tan\alpha+cot\alpha\right)=sin\alpha.cos\alpha\left(\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}\right)\)
\(=sin^2\alpha+cos^2\alpha=1\)
b) ta có : \(\left(sin^2\alpha+cos^2\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1^2+1-2sin\alpha.cos=2\left(1-2sin\alpha.cos\alpha\right)\)
c) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, = \(\sin^2\alpha+2\sin\alpha.\cos\alpha+\cos^2\alpha\)+ \(\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha\)
= \(2\sin^2\alpha+2\cos^2\alpha\)= 4
b,=\(\sin\alpha\cos\alpha\)(\(\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}\))
= \(\sin\alpha\cos\alpha.\frac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha}\)
=1
#mã mã#
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=sin^2\alpha+2sin\alpha\cdot cos\alpha+cos^2\alpha+sin^2\alpha-2sin\alpha\cdot cos\alpha+cos^2\alpha\)
\(=2\left(sin^2\alpha+cos^2\alpha\right)\)
\(=2\)
b) Vẽ hình minh họa cho dễ nhìn nè :
\(sin\alpha\cdot cos\alpha\cdot\left(tan\alpha+cot\alpha\right)\)
\(=\frac{AC}{BC}\cdot\frac{AB}{BC}\cdot\left(\frac{AC}{AB}+\frac{AB}{AC}\right)\)
\(=\frac{AC\cdot AB\cdot AC}{BC\cdot BC\cdot AB}+\frac{AC\cdot AB\cdot AB}{BC\cdot BC\cdot AC}\)
\(=\left(\frac{AC}{BC}\right)^2+\left(\frac{AB}{BC}\right)^2\)
\(=sin^2\text{α}+cos^2\text{α}\)
\(=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=\(\left(1+\frac{sin^2a}{cos^2a}\right)\).\(cos^2a\)+\(\left(1+\frac{cos^2a}{sin^2a}\right)\).\(sin^2a\)
=\(cos^2a+sin^2a+sin^2a+cos^2a\)
=\(2\left(sin^2a+cos^2a\right)\)=\(2.1=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1+\dfrac{sin^2a}{cos^2a}\right).cos^2a-\left(1+\dfrac{cos^2a}{sin^2a}\right).sin^2a\)
\(=\dfrac{\left(sin^2a+cos^2a\right)}{cos^2a}.cos^2a-\left(\dfrac{sin^2a+cos^2a}{sin^2a}\right).sin^2a\)
\(=1-1=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) khai triển được 2sin2+2cos2=2(sin2+cos2=2.1=2
b)cot2-cos2.cot2=cot2(1-cos2)=cot2.sin2=cos2/sin2.sin2=cos2
c)sin.cos(tan+cot)=sin.cos.tan+sin.cos.cot=sin.cos.sin/cos+sin.cos.cos/sin=sin2+cos2=1
d)tan2-tan2.sin2=tan2(1-sin2)=tan2.cos2=sin2/cos2.cos2=sin2
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình thay \(\alpha\) thành x để tiện ghi nhé
a) \(sinx.cosx\left(tanx+cotx\right)\)
\(=sinx.cosx\left(\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}\right)\)
\(=sinx.cosx\left(\dfrac{sinx^2+cosx^2}{sinx.cosx}\right)\)
\(=\dfrac{sinx.cosx}{sinx.cosx}=1\)
b) \(cot^2-cos^2.cot^2\)
\(=\dfrac{cos^2}{sin^2}-\left(1-sin^2\right).\dfrac{cos^2}{sin^2}\)
\(=\dfrac{cos^2-cos^2+sin^2cos^2}{sin^2}\)
\(=\dfrac{sin^2.cos^2}{sin^2}\)
\(=cos^2\)
c) \(tan^2-sin^2.tan^2\)
\(=tan^2\left(1-sin^2\right)\)
\(=\dfrac{sin^2}{cos^2}cos^2\)
\(=sin^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
tui rất thích lượng giác:
a) = s2 + 2s.c +c2 +s2- 2s.c + c2 =1+1=2
b) = s.c(s/c + c/s) = s.c(s2 + c2) / s.c = 1
.............................bài nào cx dễ
( k có việc j khó, chỉ sợ lòng k bền....)
e chưa nên k lm đk thông cảm cho e.
=\(\left(1+\frac{sin^2a}{cos^2a}\right)\)\(cos^2a\)+\(\left(1+\frac{cos^2a}{sin^2a}\right)\)\(sin^2a\)
=\(cos^2a\)+\(sin^2a\)+\(sin^2a\)+\(cos^2a\)
=\(2sin^2a\)+\(2cos^2a\)
=\(2\left(sin^2a+cos^2a\right)\)
=2