Question

化简:根号[(sin平方α×(1+cotα)+cos平方α×(1+tanα)]

化简:根号[(sin平方α×(1+cotα)+cos平方α×(1+tanα)]

Answer 1

(sin平方α×(1+cotα)+cos平方α×(1+tanα)
=sin²α*(1+cosα/sinα)+cos²α*(1+sinα/cosα)
=sin²α+sinαcosα+cos²α+cosαsinα
=sin²α+2sinαcosα+cos²α
=(sinα+cosα)²
所以 ：根号[(sin平方α×(1+cotα)+cos平方α×(1+tanα)]=|sinα+cosα|

Answer 2

√[(sin^2α×(1+cotα)+cos^2α×(1+tanα)]
=√(sin^2α+sin^2αcotα+cos^2α+cos^2αtanα)
=√(sin^2α+cos^2α+sin^2αcotα+cos^2αtanα)
=√(sin^2α+cos^2α+sin^2α*cos/sinα+sinαcosα)
=√(sin^2α+cos^2α+sinαcosα+sinαcosα)
=√(sin^2α+cos^2α+2sinαcosα)
=√(sinα+cosα)^2
=∣sinα+cosα∣

Answer 3

√[(sin²a(1+cota)+cos²a(1+tana)
=√(sin²a+sinacosa+cos²a+cosasina)
=√(sin²a+2sinacosa+cos²a)
=√(sina+cosa)²
=|sina+cosa|