a/ Ta có: \(tan\alpha=5\Rightarrow cot\alpha=\frac{1}{5}\) . Đề: \(\frac{sin\alpha}{sin^3\alpha+cos^3\alpha}=\frac{\frac{1}{sin^2\alpha}}{1+\frac{cos^3\alpha}{sin^3\alpha}}=\frac{1+cot^2\alpha}{1+cot^3\alpha}=\frac{1+\left(\frac{1}{5}\right)^2}{1+\left(\frac{1}{5}\right)^3}=\frac{65}{63}\)         

b/ Ta có vế trái \(=\frac{sin^2x+cos^2x+cos^2x-sin^2x+\left(sinx+sin3x\right)}{1+2sinx}=\frac{2cos^2x+2.sin2x.cosx}{1+2sinx}=\frac{2cos^2x+4.sinx.cos^2x}{1+2sinx}=\frac{2cos^2x.\left(1+2sinx\right)}{1+2sinx}=2cos^2x\) ( = vế phải)

Vì 0 < α < π/2 nên sin α > 0, cos α > 0, tan α > 0, cot α > 0.

`\pi/2 < \alpha < \pi=>\alpha` nằm ở góc phần tư thứ `2`

    `=>{(sin  \alpha > 0;cos \alpha < 0),(tan \alpha < 0; cot \alpha < 0):}`

      `->\bb D`

ta có \(sina< tana\\ cosa< cota\)

mà 2 góc 35 độ và 55 độ là hai góc phụ nhau nên \(cos35^o=sin55^o< tan55^o\)

tương tự: \(sin72^o=Cos12^o< cot12^o\)

Bạn có máy tính không?

\(A=3\cdot\left(\sin^4\left(x\right)+\cos^4\left(x\right)\right)-2\cdot\left(\sin^6\left(x\right)+\cos^6\left(x\right)\right)\)

\(=3\cdot\sin^4\left(x\right)+3\cdot\cos^4\left(x\right)-2\cdot\left(\left(\sin^2\left(x\right)\right)^3+\left(\cos^2\left(x\right)\right)^3\right)\)

\(=3\cdot\sin^4\left(x\right)+3\cdot\cos^4\left(x\right)-2\cdot\left(\left(\sin^2\left(x\right)+\cos^2\left(x\right)\right)\cdot\left(\sin^4\left(x\right)-\sin^2\left(x\right)\cdot\cos^2\left(x\right)+\cos^4\left(x\right)\right)\right)\)

\(=3\cdot\sin^4\left(x\right)+3\cdot\cos^4\left(x\right)-2\cdot\left(\sin^4\left(x\right)-\sin^2\left(x\right)\cos^2\left(x\right)+\cos^4\left(x\right)\right)\)

\(=3\sin^4\left(x\right)+3\cos^4\left(x\right)-2\sin^4\left(x\right)-2\cos^4\left(x\right)+2\sin^2\left(x\right)\cos^2\left(x\right)\)

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

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

\(=1^2\)

\(=1\)

Vậy kết quả của biểu thức không phụ thuộc vào giá trị của x (đpcm)

(chúc bạn học tốt)

\(A=cos\left(\pi+\frac{\pi}{2}-a\right)-sin\left(\pi+\frac{\pi}{2}-a\right)+cos\left(a+\frac{\pi}{2}-4\pi\right)-sin\left(a+\frac{\pi}{2}-4\pi\right)\)

\(=-cos\left(\frac{\pi}{2}-a\right)+sin\left(\frac{\pi}{2}-a\right)+cos\left(a+\frac{\pi}{2}\right)-sin\left(a+\frac{\pi}{2}\right)\)

\(=-sina+cosa-sina-cosa=-2sina\)

a)

Ta có:

\({\cos ^4}\alpha {\sin ^4}\alpha  = \left( {{{\cos }^2}\alpha  - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha  + {{\sin }^2}\alpha } \right) \\= {\cos ^2}\alpha  - {\sin ^2}\alpha = {\cos ^2}\alpha  - (1 - {\cos ^2}\alpha ) \\= {\cos ^2}\alpha  - 1 + {\cos ^2}\alpha  = 2{\cos ^2}\alpha  - 1\)

(đpcm)

b)

Ta có:

\(\frac{{{{\cos }^2}\alpha  + {{\tan }^2}\alpha  - 1}}{{{{\sin }^2}\alpha }} = \frac{{{{\cos }^2}\alpha \; + {{\tan }^2}\alpha  - {{\sin }^2}\alpha  - {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} \\= \frac{{{{\tan }^2}\alpha  - {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} - {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} \\= \frac{1}{{{{\cos }^2}\alpha }} - 1 = {\tan ^2}\alpha \)

(đpcm)