In quadrant IV, cosine is positive and sine is negative. This means that
[tex]\cos^2(x)+\sin^2(x)=1 \iff \sin^2(x)=1-\cos^2(x) \implies \sin(x)=-\sqrt{1-\cos^2(x)[/tex]
The cotangent is defined as the ratio between the cosine and sine:
[tex]\cot(x)=\dfrac{\cos(x)}{\sin(x)}=\dfrac{\cos(x)}{-\sqrt{1-\cos^2(x)}}=-\dfrac{2}{7}\iff \dfrac{\cos(x)}{\sqrt{1-\cos^2(x)}}=\dfrac{2}{7}[/tex]
So, we have the following equation:
[tex]2\sqrt{1-\cos^2(x)}=7\cos(x)[/tex]
Squaring both sides yields
[tex]4(1-\cos^2(x))=49\cos^2(x) \iff 4-4\cos^2(x)=49\cos^2(x)\iff 53\cos^2(x)=4[/tex]
The solution to this equation would be
[tex]\cos^2(x)=\dfrac{4}{53}\iff \cos^2(x)=\pm\dfrac{2}{\sqrt{53}}[/tex]
But we know that the cosine has to be positive, so we have
[tex]\cos(x)=\dfrac{2}{\sqrt{53}}[/tex]
And
[tex]\sin(x)=-\sqrt{1-\dfrac{4}{53}}=-\sqrt{\dfrac{49}{53}}=-\dfrac{7}{\sqrt{53}}[/tex]
Finally, the secant is the inverse of the cosine, so it's
[tex]\sec(x)=\dfrac{1}{\cos(x)}=\dfrac{\sqrt{53}}{2}[/tex]
