Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý

Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)

Mình sửa lại đề rồi á

\(A=\dfrac{cos^2a-sin^2a}{\dfrac{cos^2a}{sin^2a}-\dfrac{sin^2a}{cos^2a}}-cos^2a=\dfrac{cos^2a.sin^2a\left(cos^2a-sin^2a\right)}{\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)}-cos^2a\)

\(=cos^2a.sin^2a-cos^2a=cos^2a\left(sin^2a-1\right)=-cos^4a\)

\(B=\sqrt{\left(1-cos^2a\right)^2+6cos^2a+3cos^4a}+\sqrt{\left(1-sin^2a\right)^2+6sin^2a+3sin^4a}\)

\(=\sqrt{4cos^4a+4cos^2a+1}+\sqrt{4sin^4a+4sin^2a+1}\)

\(=\sqrt{\left(2cos^2a+1\right)^2}+\sqrt{\left(2sin^2a+1\right)^2}\)

\(=2\left(sin^2a+cos^2a\right)+2=4\)

D=sin(pi+x)+sinx+cot(pi-x)+tan(pi/2-x)

=-sinx+sinx-cotx+cotx=0

$\sin a.\cos a.(\tan a+\cot a)\\=\sin a.\cos a.\tan a+\sin a.\cos a.\cot a\\=\sin a.\cos a.\dfrac{\sin a}{\cos a}+\sin a.\cos a.\dfrac{\cos a}{\sin a}\\=\sin^2 a+\cos^2 a\\=1$

\(sin\left(a\right).cos\left(a\right).\left(tan\left(a\right)+cot\left(a\right)\right)\\ =sin\left(a\right).cos\left(a\right).tan\left(a\right)+sin\left(a\right).cos\left(a\right).cot\left(a\right)\\ =sin\left(a\right).cos\left(a\right).\dfrac{sin\left(a\right)}{cos\left(a\right)}+sin\left(a\right).cos\left(a\right).\dfrac{cos\left(a\right)}{sin\left(a\right)}\\ =sin^2\left(a\right)+cos^2\left(a\right)=1\)

a:Sửa đề: \(I=\frac{cos\left(-288^0\right)\cdot\cot72^0}{\tan\left(-162^0\right)\cdot\sin108^0}-\tan18^0\)

Ta có: \(I=\frac{cos\left(-288^0\right)\cdot\cot72^0}{\tan\left(-162^0\right)\cdot\sin108^0}-\tan18^0\)

\(=\frac{cos\left(72^0-360^0\right)\cdot\cot72^0}{\tan\left(18^0-180^0\right)\cdot\sin108^0}-\tan18^0\)

\(=\frac{cos72^0\cdot\cot72^0}{\tan18^0\cdot\sin108^0}-\tan18^0\)

\(=\frac{cos72^0\cdot\frac{cos72^0}{\sin72^0}}{\tan18^0\cdot\sin108^0}-\tan18^0=\frac{cos^272^0}{\sin72^0\cdot\tan18^0\cdot\sin\left(180^0-108^0\right)}-\tan18^0\)

\(=\frac{cos^272^0}{\sin72^0\cdot\tan18^0\cdot\sin72^0}-\tan18^0=\frac{cos^272^0}{\sin^272^0\cdot\tan18^0}-\tan18^0\)

\(=\frac{cos^272^0}{\sin^272^0\cdot\cot72^0}-\tan18^0=\frac{cos^272^0}{\sin^272^0\cdot\frac{cos72^0}{\sin72^0}}-\tan18^0\)

\(=\frac{cos72^0}{\sin72^0}-\tan18^0=\cot72^0-\tan18^0=0\)

b: Ta có: \(J=2\cdot\sin\left(790^0+x\right)+cos\left(1260^0-x\right)+\tan\left(630^0+x\right)\cdot\tan\left(1260^0-x\right)\)

\(=2\cdot\sin\left(720^0+70^0+x\right)+cos\left(1080^0+180^0-x\right)+\tan\left(720^0+x-90^0\right)\cdot\tan\left(1080^0+180^0-x\right)\)

\(=2\cdot\sin\left(70^0+x\right)+cos\left(180^0-x\right)+\tan\left(x-90^0\right)\cdot\tan\left(180^0-x\right)\)

\(=2\cdot\sin\left(70^0+x\right)-cosx-\tan\left(90^0-x\right)\cdot\left(-\tan x\right)\)

\(=2\cdot\sin\left(70^0+x\right)-cosx+\tan\left(90^0-x\right)\cdot\tan x\)

\(=2\cdot\sin\left(70^0+x\right)-cosx+\cot x\cdot\tan x=2\cdot\sin\left(x+70^0\right)-cosx+1\)

\(A=tan18^otan288+sin32^osin148^o-sin302^osin122^o\)
\(=tan18^o.tan\left(-72^o\right)+sin32^o.sin32^o+sin58^o.sin58^o\)
\(=-tan18^o.cot18^o+sin^232^o+sin^258^o\)
\(=-1+sin^232^o+cos^232^2=-1+1=0\).

b) \(B=\dfrac{1+sin^4\alpha-cos^4\alpha}{1-sin^6\alpha-cos^6\alpha}\)
\(=\dfrac{1+\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)}{1-\left(sin^6\alpha+cos^6\alpha\right)}\)
\(=\dfrac{1+sin^2\alpha-cos^2\alpha}{1-\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha\right)}\)
\(=\dfrac{sin^2\alpha+1-cos^2\alpha}{1-\left(1-sin\alpha.cos\alpha\right)}\)
\(=\dfrac{sin^2\alpha+sin^2\alpha}{sin\alpha cos\alpha}\)
\(=\dfrac{2sin^2\alpha}{sin\alpha cos\alpha}=\dfrac{2sin\alpha}{cos\alpha}=2tan\alpha\).