Albert wants to show that tan(theta)sin(theta)+cos(theta)=sec (theta). he writes the following proof: tan(theta)sin(theta)+cos(theta)=sec (theta) sin(theta)/cos(theta) sin(theta)+cos(theta)=sec(theta) sin^2(theta)/cos(theta)+cos(theta)= sec(theta) what is the next step in this proof? a.) he should write tan(theta)=sin(theta)/cos(theta) to find a common denominator. b.) he should write cos(theta)=cos^2(theta)/cos(theta) to find a common denominator. c.) he should write cos(theta)=1-sin(theta) to convert all the terms to sine. d.) he should write sin(theta)=1-cos(theta) to convert all the terms to cosine.

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.

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