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Đáp án:

Giải thích các bước giải:

 Câu 41 

`a) sin^2 alpha  + cos^2 alpha =1 ` 

`<=>cos^2 alpha = 1-(5/13)^2`

`<=> cos^2 alpha =144/169`

`<=> cos alpha = sqrt(144/169)`

Mà `90^o<alpha <180^o`

`=> cos alpha = -12/13`

`tan alpha = (sin alpha )/ (cos alpha ) = (5/13)/(-12/13) = -5/12` 

`b)  cos alpha = -3/5`

`<=>sin^2 alpha +cos^2alpha = 1`

`<=> sin^2 alpha  =1- (-3/5)^2`

`<=> sin ^2 alpha  = 16/25`

`<=> sin alpha =4/5 ` ( vì `sin alpha ` luôn dương)

`cot alpha  = (cos alpha )/ (sin alpha ) = (-3/5)/(4/5)= -3/4`

c) `tan beta = -7/24 => 90^o <beta<180^o `

`cot beta= 1/(tan beta ) = -24/7`

` 1+ tan^2 beta=1 / ( cos^2 beta)`

`<=> cos^2 beta = 1/ (1 + (-7/24)^2) `

`<=> cos^2 beta=576/625`

`<=> cos beta= 24/25`

` 1+ cot^2 beta=1 / (sin^2 beta)`

`<=> sin^2 beta = 1/ (1 + (-24/7)^2) `

`<=> sin^2 beta=49/625`

`<=> sin beta= 7/25`