Given the equation:
[tex]2tan^2x-secx=1[/tex]Let's find the true equations
First simplify using trigonometric identity
[tex]2(sec^2x-1)-secx=1[/tex]Apply distributive property:
[tex]\begin{gathered} 2sec^2x-2-secx=1 \\ \\ Add\text{ 2 to both sides:} \\ 2sec^2x-2+2-secx=1+2 \\ \\ 2sec^2x-secx=3 \end{gathered}[/tex]• Now, let's verify where: sec x = -1:
[tex]\begin{gathered} 2(-1)^2-(-1)=3 \\ \\ 2+1=3 \\ \\ 3=3 \end{gathered}[/tex]Hence, secx = 1 is true.
• For sec x = 3:
[tex]\begin{gathered} 2(3)^2-3^2=3 \\ \\ 18-9=3 \\ \\ 9=3 \end{gathered}[/tex]Hence, sec(x) = 3 is not true,
• For tanx = 3, we have:
Rewrite using trigonometric identity
[tex]\begin{gathered} 2tan^2x-\sqrt{1+tan^2x}=1 \\ \\ 2(3)^2-\sqrt{1+(3)^2}=1 \\ \\ 18-\sqrt{10}=1 \end{gathered}[/tex]tan x = 3 is not true,
• For tanx = -1:
[tex]\begin{gathered} 2tan^2x-\sqrt{1+tan^2x}=1 \\ \\ 2(-1)^2-\sqrt{1+(-1)^2}=1 \\ \\ -2-\sqrt{2}=1 \end{gathered}[/tex]Hence tanx = -1 is not true.
• Now, for sec x = 3/2, we have:
[tex]\begin{gathered} 2sec^2x-secx=3 \\ \\ 2(\frac{3}{2})^2-(\frac{3}{2})=3 \\ \\ 2(\frac{9}{4})-\frac{3}{2}=3 \\ \\ \frac{9}{2}-\frac{3}{2}=3 \\ \\ 3=3 \end{gathered}[/tex]Hence, secx = 3/2 is true.
Therefore, the true equations are:
A. secx = -1
E. secx = 3/2
ANSWER:
A. sec x = -1
E. sec x = 3/2
