11.16 repeat only the six is repeating
We have that
<span>tan(theta)sin(theta)+cos(theta)=sec(theta)
</span><span>[sin(theta)/cos(theta)] sin(theta)+cos(theta)=sec(theta)
</span>[sin²<span>(theta)/cos(theta)]+cos(theta)=sec(theta)
</span><span>the next step in this proof
is </span>write cos(theta)=cos²<span>(theta)/cos(theta) to find a common denominator
so
</span>[sin²(theta)/cos(theta)]+[cos²(theta)/cos(theta)]=sec(theta)<span>
</span>{[sin²(theta)+cos²(theta)]/cos(theta)}=sec(theta)<span>
remember that
</span>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 <span>B.)
He should write cos(theta)=cos^2(theta)/cos(theta) to find a common denominator.</span>
k and m must satisfy 8k = 1 mean k = ⅛
and 2km+⅝=0 mean m = -5/2
