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\(\frac{cos^2x\left(1+cot^2x\right)}{sin^2x\left(1+tan^2x\right)}=\frac{tan^2x\left(1+cot^2x\right)}{1+tan^2x}=\frac{tan^2x+tan^2x.cot^2x}{1+tan^2x}=\frac{1+tan^2x}{1+tan^2x}=1\)
Câu b ko rút gọn được, bạn coi lại đề
\(x^2sin^2a+y^2cos^2a-2xy.sina.cosa+x^2cos^2a+y^2sin^2a+2xy.sinx.cosa\)
\(=x^2\left(sin^2a+cos^2a\right)+y^2\left(cos^2a+sin^2a\right)=x^2+y^2\)
a) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{3}{4}=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{cases}}\)
- \(cosx=\frac{1}{2}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)
- \(cosx=\frac{-1}{2}\):
\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=-\sqrt{3}\)
\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\sqrt{3}}=\frac{-\sqrt{3}}{3}\)
b) Bạn làm tương tự câu a) nha.
a) \(cos^4x-sin^4x=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=cos^2x-sin^2x\)
b) \(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{tanxcotx}{tanxcotx+cotx}=\frac{1}{1+tanx}+\frac{tanx}{tanx+1}\)
\(=\frac{1+tanx}{1+tanx}=1\)
c) Ta có: \(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}\)
\(\Rightarrow\frac{1}{1+tan^2x}=cos^2x\)
Tương tự \(\frac{1}{1+tan^2y}=cos^2y\)
\(\Rightarrow cos^2x-cos^2y=\frac{1}{1+tan^2x}-\frac{1}{1+tan^2y}\)
\(cos^2x-cos^2y=\left(1-sin^2x\right)-\left(1-sin^2y\right)=sin^2y-sin^2x\)
d) \(\frac{1+sin^2x}{1-sin^2x}=\frac{cos^2x+sin^2x+sin^2x}{cos^2x+sin^2x-sin^2x}=\frac{cos^2x+2sin^2x}{cos^2x}=1+2\left(\frac{sinx}{cosx}\right)^2=1+2tan^2x\)
Xét tam giác ABC vuông tại A có AH là đường cao và AM là trung tuyến
Đặt \(\widehat{MAC}=\widehat{MCA}=x\)thì \(\widehat{BMA}=2x\)(theo tính chất đường trung tuyến ứng với cạnh huyền của tam giác vuông)
a) Ta có: \(\sin2x=\frac{AH}{AM}=2.\frac{AH}{BC}=2.\frac{AH}{AC}.\frac{AC}{BC}=2.\sin ACH.\cos ACB=2\cos x.\sin x\)
b) \(\cos2x=\frac{HM}{AM}=\frac{2HM}{BC}=\frac{2HC-2CM}{BC}=2.\frac{HC}{BC}-1=2.\frac{HC}{ AC}.\frac{AC}{BC}-1=2.\cos ACH.\cos ACB-1=2\cos^2x-1=2\cos^2x-\left(\sin^2x+\cos^2x\right)=\cos^2x-\sin^2x\)c) \(\tan2x=\frac{\sin2x}{\cos2x}=\frac{2\cos x.\sin x}{\cos^2x-\sin^2x}=\frac{2.\frac{\sin x}{\cos x}}{\frac{\cos^2x}{\cos^2x}-\frac{\sin^2x}{\cos^2x}}=\frac{2\tan x}{1-\tan^2x}\)
a) \(\sqrt{\frac{1+\cos x}{1-\cos x}}-\sqrt{\frac{1-\cos x}{1+\cos x}}=\frac{\sqrt{\left(1+\cos x\right)^2}-\sqrt{\left(1-\cos x\right)^2}}{\sqrt{\left(1-\cos x\right)\left(1+\cos x\right)}}\)
\(=\frac{1+\cos x-1+\cos x}{\sqrt{1-\cos^2x}}=\frac{2\cos x}{\sqrt{\sin^2x}}=\frac{2\cos x}{\sin x}=2\cot x\)
b) \(\frac{1}{\tan x+1}+\frac{1}{\cot x+1}=\frac{\tan x+1+\cot x+1}{\left(\tan x+1\right)\left(\cot x+1\right)}\)
\(=\frac{\tan x+\cot x+2}{\tan x+\cot x+\tan x.\cot x+1}=\frac{\tan x+\cot x+2}{\tan x+\cot x+2}=1\)
c) (ko bt có sai đề ko, làm mãi ko ra)
d) \(\sin^21^0+\sin^22^0+\sin^23^0+...+\sin^289^0\)
\(=\left(\sin^21^0+\sin^289^0\right)+\left(\sin^22^0+\sin^288^0\right)+...+\sin^245^0\)
\(=\left[\left(\sin^21^0-\cos^289^0\right)+\left(\sin^289^0+\cos^289^0\right)\right]+\)
\(\left[\left(\sin^22^0-\cos^288^0\right)+\left(\sin^288^0+\cos^288^0\right)\right]+...+\sin^245^0\)
\(=\left(0+1\right)+\left(0+1\right)+...+\frac{\sqrt{2}}{2}=\frac{44+\sqrt{2}}{2}\)