Respuesta :
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.
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.
He should write cos(theta)=cos^2(theta)/cos(theta) to find a common denominator.
We have given 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)
B) 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) .........(1)
What is the value of sin²(theta)+cos²(theta)?
sin²(theta)+cos²(theta)=1
So, left side of equation 1 is,
={[sin²(theta)+cos²(theta)]/cos(theta)}
= 1/cos(theta)
=sec(theta)
We get
Left side=right side
Hence the proof.
That is the answer B is correct.
Therefor,
He should write cos(theta)=cos^2(theta)/cos(theta) to find a common denominator.
To learn more about trigonometric function visit:
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