a) \(\dfrac{1}{\tan\alpha+1}+\dfrac{1}{\cot\alpha+1}\) \(=\dfrac{\tan\alpha+1+\cot\alpha+1}{\left(\tan\alpha+1\right)\left(\cot\alpha+1\right)}\) \(=\dfrac{\tan\alpha+\cot\alpha+2}{\tan\alpha\cot\alpha+\tan\alpha+\cot\alpha+1}\) \(=1\) (vì \(\tan\alpha\cot\alpha=1\))

b) \(\cos\left(\dfrac{\pi}{2}-\alpha\right)-\sin\left(\pi+\alpha\right)\) \(=\sin\left(\alpha\right)-\sin\left(\pi-\alpha\right)\) \(=0\) (do \(\sin\) của 2 cung bù nhau thì bằng nhau, \(\cos\) của 1 góc bằng \(\sin\) của góc phụ với nó).

c) \(\sin\left(\alpha-\dfrac{\pi}{2}\right)+\cos\left(-\alpha+6\pi\right)-\tan\left(\alpha+\pi\right)\cot\left(3\pi-\alpha\right)\)

\(=\cos\left(\pi-\alpha\right)+\cos\left(-\alpha\right)-\tan\alpha\cot\left(\pi-\alpha\right)\)

\(=\tan\alpha\cot\alpha\) \(=1\) (ở đây áp dụng tính chất của 2 cung hơn kém \(\pi\) nhiều lần)

Có \(\sin a-\cos a=-\sqrt{2}\left(-\sin a.\sin\dfrac{\pi}{4}+\cos a.\cos\dfrac{\pi}{4}\right)\)

\(=-\sqrt{2}\cos\left(a+\dfrac{\pi}{4}\right)\)

\(\Rightarrow\left(\sin a-\cos a\right)^2=2.\cos^2\left(a+\dfrac{\pi}{4}\right)\)

\(VT=sin^4x\cdot\dfrac{cos^2x}{sin^2x}+cos^4x\cdot\dfrac{sin^2x}{cos^2x}+sin^4x-sin^2x\cdot cos^2x\)

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

\(=sin^2x\left(sin^2x+cos^2x\right)=sin^2x=VP\)

c: 2(sin^6a+cos^6a)+1

=2[(sin^2a+cos^2a)^3-3*sin^2a*cos^2a]+1

=2-6sin^2acos^2a+1

=3-6*sin^2a*cos^2a

=3(sin^4a+cos^4a)

a:

Sửa đề: =-tana*tanb

 \(VT=\left(\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}\right):\left(\dfrac{cosa}{sina}-\dfrac{cosb}{sinb}\right)\)

\(=\dfrac{sina\cdot cosb-sinb\cdot cosa}{cosa\cdot cosb}:\dfrac{cosa\cdot sinb-cosb\cdot sina}{sina\cdot sinb}\)

\(=\dfrac{sin\left(a-b\right)}{cosa\cdot cosb}\cdot\dfrac{sina\cdot sinb}{sin\left(b-a\right)}\)

\(=-tana\cdot tanb\)

=VP

Ta có:

`sin^4 \alpha + cos^4 \alpha -sin^6 \alpha- cos^6\alpha`

`=sin^4\alpha+cos^4\alpha-(sin^2\alpha+cos^2\alpha)(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha)`

`=sin^4\alpha + cos^4\alpha-(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha)`

`=sin^2\alpha cos^2\alpha(ĐPCM)`

a: \(VT=\dfrac{\left(sina+cosa\right)^3-3\cdot sina\cdot cosa\left(sina+cosa\right)}{sina+cosa}\)

=(sina+cosa)^2-3*sina*cosa

=sin^2a+cos^2a-sina*cosa

=1-sina*cosa=VP

c: VT=(sin^2a+cos^2a)^2-2*sin^2a*cos^2a-(sin^2a+cos^2a)^3+3*sin^2a*cos^2a*(sin^2a+cos^2a)

=1-2sin^2a*cos^2a-1+3*sin^2a*cos^2a

=sin^2a*cos^2a=VP

\(a,cos^4a-sin^4a=2cos^2a-1\\ VT=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)\\ =cos^2a-sin^2a\\ =cos2a=2cos^2a-1\)

\(b,VT=\dfrac{cos^2a+\dfrac{sin^2a}{cos^2a}-1}{sin^2a}\\ =\dfrac{\dfrac{cos^4a+sin^2a-cos^2a}{cos^2a}}{sin^2a}\\ =\dfrac{\dfrac{cos^4a+\left(1-cos^2a\right)-cos^2a}{cos^2a}}{sin^2a}\\ =\dfrac{\dfrac{cos^4a+1-2cos^2a}{cos^2a}}{sin^2a}\\ =\dfrac{\dfrac{\left(1-cos^2a\right)^2}{cos^2a}}{sin^2a}\\ =\dfrac{sin^4a}{cos^2a}:sin^2a\\ =\dfrac{sin^4a}{cos^2a}\times\dfrac{1}{sin^2a}\\ =\dfrac{sin^2a}{cos^2a}=tan^2a\)

Tự vẽ hình nhé!

Ta có:

\(V_{OBCNM}=\dfrac{1}{3}d\left(O;\left(BCNM\right)\right).S_{BCNM}=\dfrac{1}{3}.\dfrac{1}{2}d\left(A;\left(SBC\right)\right).\dfrac{3}{4}S_{SBC}=\dfrac{1}{8}V_{SABC}=\dfrac{1}{16}V_{SABCD}\)

\(\Rightarrow\dfrac{V_{OBCNM}}{V_{SABCD}}=\dfrac{1}{16}\)