`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`

Chọn A.

Chọn C.

Ta có tan α – cotα = 1 

Do  suy ra tanα < 0 nên 

Thay

 và

vào P  ta được 

Chọn B.

Ta có: 1 + cos2α = 2cos²α và sin2α = 2sinα.cosα.

Mà tanα = 2 nên cot α = 1/2

Suy ra:

Chọn D.

Chọn B.

Theo giả thiết:

Ta có:

\(F=cos\left(\frac{\pi}{4}+a\right)\cdot cos\left(\frac{\pi}{4}-a\right)\)

\(=\frac12\cdot\left\lbrack cos\left(\frac{\pi}{4}+a-\frac{\pi}{4}+a\right)+cos\left(\frac{\pi}{4}+a+\frac{\pi}{4}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack cos\left(2a\right)+cos\left(\frac{\pi}{2}\right)\right\rbrack=\frac12\cdot cos2a\)

\(G=\sin\left(\frac{\pi}{3}+a\right)\cdot cos\left(\frac{\pi}{3}-a\right)\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{3}+a+\frac{\pi}{3}-a\right)+\sin\left(\frac{\pi}{3}+a-\frac{\pi}{3}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac23\pi\right)+\sin2a\right\rbrack=\frac12\cdot\left\lbrack\frac12+\sin2a\right\rbrack\)

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

\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{2}+a+\frac{\pi}{2}-a\right)+\sin\left(\frac{\pi}{2}+a-\frac{\pi}{2}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack\sin\left(\pi\right)+\sin2a\right\rbrack=\frac12\left\lbrack2\cdot\sin a\cdot cosa\right\rbrack=\sin a\cdot cosa\)

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

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

=0

\(K=cos\left(\frac{\pi}{6}-x\right)-\sin\left(\frac{\pi}{3}+x\right)\)

\(=\sin\left(\frac{\pi}{2}-\frac{\pi}{6}+x\right)-\sin\left(\frac{\pi}{3}+x\right)=\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}+x\right)\)

=0