Question

高一数学三角函数化简为题

√[(sin^2 α)(1+cotα)+(cos^2 α)(1+tanα)]

Answer 1

解答：
√[(sin^2 α)(1+cotα)+(cos^2 α)(1+tanα)]
=√[(sin)^2+sinacosa+sinacosa+(cosa)^2]
=√(sina+cosa)^2
=|sina+cosa|

Answer 2

sin^2α/（1+cotα)+cos^2α/（1+tanα)
=sin^2α·sinα/（sinα+cosα)+cos^2α·cosα/（cosα+sinα)
=(sin^3α+cos^3α)/（sinα+cosα)
=(sinα+cosα)(sin^2α-sinα·cosα+cos^2α)/（sinα+cosα)
=sin^2α-sinα·cosα+cos^2α
=1-sinαcosα

Answer 3

将切化成弦
√[(sin^2 α)(1+cotα)+(cos^2 α)(1+tanα)]
=√[sinα(sinα+cosα)+cosα(cosα+sinα)]
=√(sinα+cosα)^2
=|sinα+cosα|