K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
LV
1
CH
Cô Hoàng Huyền
Admin
VIP
26 tháng 9 2017
\(S=\frac{cos^2a-sin^2b}{sin^2a.sin^2b}-cot^2a.cot^2b=\frac{cos^2a-sin^2b}{sin^2a.sin^2b}-\frac{cos^2a.cos^2b}{sin^2a.sin^2b}\)
\(=\frac{cos^2a-sin^2b-cos^2a.cos^2b}{sin^2a.sin^2b}=\frac{cos^2a-cos^2a.cos^2b-sin^2b}{sin^2a.sin^2b}\)
\(=\frac{cos^2a\left(1-cos^2b\right)-sin^2b}{sin^2a.sin^2b}=\frac{cos^2a.sin^2b-sin^2b}{sin^2a.sin^2b}\)
\(=\frac{sin^2b\left(cos^2a-1\right)}{sin^2a.sin^2b}=\frac{-sin^2a.sin^2b}{sin^2a.sin^2b}=-1.\)
MR
1
23 tháng 5 2022
\(P=\dfrac{\sin60^0+\tan^230^0}{\cos30^0-\cot60^0}=\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{3}\right):\left(\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{3}\right)\)
\(=\dfrac{3\sqrt{3}+2}{6}:\dfrac{3\sqrt{3}-2\sqrt{3}}{6}\)
\(=\dfrac{3\sqrt{3}+2}{\sqrt{3}}=\dfrac{6+2\sqrt{3}}{3}\)
KN
30 tháng 10 2020
a) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos^2a=1-\sin^2\alpha=1-\left(\frac{\sqrt{3}}{2}\right)^2=\frac{1}{4}\)
\(\Rightarrow\cos\alpha=\frac{1}{2}\)(do \(\cos\alpha>0\))
b) \(Q=\sin^2\alpha+\cot^2\alpha.\sin^2\alpha=\sin^2\alpha\left(1+\cot^2\alpha\right)\)\(=\sin^2\alpha\left(1+\frac{\cos^2\alpha}{\sin^2\alpha}\right)=\sin^2\alpha.\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}=1\)
KN
30 tháng 10 2020
a) \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)
CN
2
5 tháng 6 2020
\(sin^2a+cos^2a-sin^4a-2cos^2a+sin^2a\)
\(=2sin^2a-cos^2a-sin^4a\)
\(=2sin^2a-cos^2a-\left(\frac{1-cos2a}{2}\right)^2\)
khai triển ra rồi quy đồng lên
\(=\frac{8sin^2a-4cos^2a-1+2cos2a-cos^22a}{4}\)
Mà \(2cos2a=2\left(cos^2a-1\right)=4cos^2-2\)
\(\Rightarrow\frac{8sin^2a-cos^22a-3}{4}\)
Mà \(-cos^22a=sin^22a-1=4sin^2cos^2-1\)
\(\Rightarrow\frac{8sin^2a+4sin^2a.cos^2a-4}{4}\)
\(=\frac{4sin^2a.\left(2-cos^2a\right)-4}{4}\)
\(=sin^2a\left(1+sin^2a\right)-1\)
\(=sin^4a-cos^2a\)
DT
9 tháng 9 2018
\(\left(1+\frac{\sin^2}{\cos^2}\right)cos^2-\left(1+\frac{cos^2}{sin^2}\right)sin^2.\)
=> \(\frac{cos^2+sin^2}{cos^2}\left(cos^2\right)-\frac{sin^2+cos^2}{sin^2}\left(sin^2\right)\)
=> 1-1 =0
24 tháng 9 2020
\(=\frac{1}{cos^2a}\cdot cos^2a+\frac{1}{sin^2a}\cdot sin^2a\)
\(=1+1\)
\(=2\)
24 tháng 8 2017
a. \(\dfrac{1+2sin\alpha cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{sin^2\alpha+2sin\alpha cos\alpha+cos^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{\left(sin\alpha+cos\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{sin\alpha+cos\alpha}{cos\alpha-sin\alpha}\)
24 tháng 5 2022
Bài 1:
\(=\left(\sin20^0-\cos70^0\right)+\left(-\tan40^0+\cot50^0\right)=0+0=0\)
Bài 2:
\(\cos a=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}\)
\(A=2\cdot\sin^2a+3\cdot\cos^2a=2\cdot\dfrac{4}{9}+3\cdot\dfrac{5}{9}=\dfrac{8+15}{9}=\dfrac{23}{9}\)
16 tháng 7 2021
Xét ΔBAC vuông tại B có a = ^A ta có :
a) \(\frac{\sin\alpha}{\cos\alpha}=\frac{\sin A}{\cos A}=\frac{\frac{BC}{AB}}{\frac{AB}{AC}}=\frac{BC}{AB}\cdot\frac{AC}{AB}=\frac{BC}{AB}=\tan A=\tan\alpha\left(đpcm\right)\)
b) \(\frac{\cos\alpha}{\sin\alpha}=\frac{\cos A}{\sin A}=\frac{\frac{AB}{AC}}{\frac{BC}{AC}}=\frac{AB}{AC}\cdot\frac{AC}{BC}=\frac{AB}{BC}=\cot A=\cot\alpha\left(đpcm\right)\)
c) \(\tan\alpha\cdot\cot\alpha=\tan A\cdot\cot A=\frac{BC}{AB}\cdot\frac{AB}{BC}=1\left(đpcm\right)\)
d) \(\sin^2\alpha+\cos^2\alpha=\sin^2A+\cos^2A=\frac{BC^2}{AC^2}+\frac{AB^2}{AC^2}=\frac{AB^2+BC^2}{AC^2}=1\left(đpcm\right)\)
e) \(\frac{1}{\cos^2\alpha}=\frac{1}{\cos^2A}=\frac{1}{\frac{AB^2}{AC^2}}=\frac{AC^2}{AB^2};1+\tan^2\alpha=1+\tan^2A=1+\frac{BC^2}{AB^2}=\frac{AB^2+BC^2}{AB^2}=\frac{AC^2}{AB^2}\)
\(\Rightarrow1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\left(đpcm\right)\)
f) \(\frac{1}{\sin^2\alpha}=\frac{1}{\sin^2A}=\frac{1}{\frac{BC^2}{AC^2}}=\frac{AC^2}{BC^2};1+\cot^2\alpha=1+\cot^2A=1+\frac{AB^2}{BC^2}=\frac{BC^2+AB^2}{BC^2}=\frac{AC^2}{BC^2}\)
\(\Rightarrow1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\left(đpcm\right)\)
\(=sin^2a\left(1+cot^2a\right)=sin^2a\left(1+\frac{cos^2a}{sin^2a}\right)=sin^2a\left(\frac{sin^2a+cos^2a}{sin^2a}\right)=\frac{sin^2a}{sin^2a}=1\)
cảm ơn bạn nhiều ạ