若tanα=3,则(sinα+cosα)²=

要有过程

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tanα=3>0得sina、cosa同号，所以sina/cosa=3,得sin^2 a=3/4,cos^2 a=1/搭汪4,
所以sinacosa=(根知野仔号3)/4(sinα+cosα)²脊镇=1+(根号3)/2=(2+根号3）/2

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sinα/cosα=3，sin²α+cos²α=1.
解得，物绝cosα=√10/戚猜10,sinα=3√10/10.
(sinα+cosα)²
=1+2sinαcosα
=1+2×√10/10×3√10/罩仔姿10
=8/5.

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答：
tana=3
(sina+cosa)^2
=(sina)^2+2sinacosa+(cosa)^2
=1+2tana*cosa*cosa
=1+2*3*(cosa)^2
=1+6/磨告[1/(cosa)^2]
=1+6/[1+(tana)^2]
=1+6/瞎圆明(1+3^2)
=1+6/10
=8/腔大5

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(sinα+cosα)²塌伏
=(sin^2α+cos^2α+2sinαcosα)/1
=(sin^2α+cos^2α+2sinαcosα)/(sin^2α+cos^2α)
上首银下同时除以cos^2a
=(tan^2α+1+2tanα)/(tan^2α+1)
=(9+1+6)/团芹携(9+1)
=1.6

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解
(sina+cosa)²——除以cos²a+sin²a=1，值不变
=(sin²a+2sinacosa+cos²a)/哗正(sin²a+cos²a)——分子分母同时除以辩卜cos²a
=(tan²a+2tana+1)/携芦穗(tan²a+1)
=(9+6+1)/(9+1)
=16/10
=8/5

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