Câu `3:`

Ta có: `sin^2 alpha=1-cos^2 alpha=(1-cos alpha)(1+cos alpha)`

`A=(1+ cos alpha)/(sin alpha) [1-(1-cos alpha)^2/(sin^2 alpha)]`

`A=(1+ cos alpha)/(sin alpha) . (sin^2 alpha-(1-cos alpha)^2)/(sin^2 alpha)`

`A= (1+ cos alpha)/(sin alpha) . ((1-cos^2 alpha)-(1-2 cos alpha + cos^2 alpha))/(sin^2 alpha)`

`A=(1+cos alpha)/(sin alpha) . (2 cos alpha-2cos^2 alpha)/(sin^2 alpha)`

`A=(1+cos alpha)/(sin alpha) . (2 cos alpha(1-cos alpha))/(sin^2 alpha)`

`A=(2 cos alpha(1-cos^2 alpha))/(sin^3 alpha)`

`A=(2 cos alpha . sin^2 alpha)/(sin^3 alpha)`

`A=(2 cos alpha)/(sin alpha)`

`A=2 cot alpha`

Theo bài ra, `tan alpha=2`

Mà `cot alpha=1/(tan alpha)=1/2`

`=> A=2 . 1/2=1`

Vậy `A=bb1`

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