Albert wants to show that tan(theta)sin(theta)+cos(theta)=sec(theta). He writes the following proof:

2 answers:

6 0

We have that
tan(theta)sin(theta)+cos(theta)=sec(theta)
[sin(theta)/cos(theta)] sin(theta)+cos(theta)=sec(theta)
[sin²(theta)/cos(theta)]+cos(theta)=sec(theta)

the next step in this proof
is write cos(theta)=cos²(theta)/cos(theta) to find a common denominator
so

[sin²(theta)/cos(theta)]+[cos²(theta)/cos(theta)]=sec(theta)

{[sin²(theta)+cos²(theta)]/cos(theta)}=sec(theta)

remember that
sin²(theta)+cos²(theta)=1
{[sin²(theta)+cos²(theta)]/cos(theta)}------------> 1/cos(theta)
and
1/cos(theta)=sec(theta)-------------> is ok

the answer is the option B.)
He should write cos(theta)=cos^2(theta)/cos(theta) to find a common denominator.

Svetlanka [38]3 years ago

4 0

Answer:

1. B, E

2. B, E

3. B

4. B

5. D

6. A

7. D

8. A

9. B

10. B

11. A

Step-by-step explanation:

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