Question

In a right angled $\Delta ABC$ right angled at $B$ the ratio of $AB$ to $AC$ is $1:\sqrt{5}$ then $3\tan \theta +5{\sec }^2\theta$ is

• A. $\frac{2}{\sqrt{5}}$
• B. $3+\sqrt{5}$
• C. $\frac{25}{4}+\frac{3}{2}$
• D. $\frac{\sqrt{5+1}}{2}$

Answer 1

The correct option is C $\frac{25}{4}+\frac{3}{2}$

$h=\sqrt{5}x$

$p=x$

$b=\sqrt{(\sqrt{5}{)}^2-x^2}=\sqrt{4x^2}=2x$

$\tan \theta =\frac{p}{b}=\frac{1x}{2x}=\frac{1}{2}$

$\sec \theta =\frac{h}{b}=\frac{\sqrt{5}x}{2x}=\frac{\sqrt{5}}{2}$

$3\tan \theta +5{\sec }^2\theta =3\times \frac{1}{2}+5{\times \left(\frac{\sqrt{5}}{2}\right)}^2=\frac{3}{2}+\frac{25}{4}$