Question

两道函数题目：tan²a-sin²a-tan²asin²a=？谢谢

tan²a-sin²a-tan²asin²a=？
若tana=-1/2,则5cos²a+3sinacosa-2sin²a=？

Answer 1

解1 tan²a-sin²a-tan²asin²a

=tan²a-tan²asin²a-sin²a

=tan²a（1-sin²a）-sin²a
=tan²a（cos²a）-sin²a
=[sin²a/cos²a]（cos²a）-sin²a
=sin²a-sin²a
=0
2 5cos²a+3sinacosa-2sin²a

=(5cos²a+3sinacosa-2sin²a)/1

=(5cos²a+3sinacosa-2sin²a)/(cos²a+sin²a)
=[(5cos²a+3sinacosa-2sin²a)×1/cos²a]/[(cos²a+sin²a)×1/cos²a]
=[(5+3tana-2tan²a)]/[(1+tan²a）
=[(5+3（-1/2)-2(-1/2)²)]/[(1+(-1/2)²）
=[5-3/2-1/2]/[(1+1/4）
=[5-2]/[(1+1/4）
=3/(5/4)
=12/5

Answer 2

答：
1）
tan²a-sin²a-tan²asin²a
=tan²a(1-sin²a)-sin²a

=tan²acos²a-sin²a
=sin²a-sin²a
=0

2)
tana=-1/2

5cos²a+3sinacosa-2sin²a

=(5+3tana-2tan²a)cos²a
=(5-3/2-2/4)cos²a
=3/(1/cos²a)
=3/[(sin²a+cos²a)/cos²a]
=3/(tan²a+1)
=3/(1/4+1)
=12/5

Answer 3

1) ,
tan²a-sin²a-tan²asin²a
=tan²a(1-sin²a)-sin²a
=tan²acos²a- sin²a
=sin²a-sin²a
=0

2),
tana=-1/2, sina=-1/√5, cosa=2/√5
5cos²a+3sinacosa-2sin²a=5*(2/√5)^2-3*/√5)*(2/√5)-2*(1/√5)^2=4-6/5-2/5=12/5