Question

If $2\sec \theta +3\tan \theta =7(0^{\circ }<\theta <{90}^{\circ })$, then the value of
$2tan\theta +3\sec \theta$ is

• A. $2\sqrt{3}$
• B. $3\sqrt{6}$
• C. $7$
• D. $\frac{1}{7}$

Answer 1

The correct option is B $3\sqrt{6}$

Let $2\tan \theta +3\sec \theta =x$

$\therefore$

$2\sec \theta +3\tan \theta =7$ .......$\left(1\right)$

$3\sec \theta +2\tan \theta =x$ .......$\left(2\right)$

$3\times$eqn$\left(1\right)-2\times$eqn$\left(2\right)$ we get

$\Rightarrow 6\sec \theta +9\tan \theta -6\sec \theta -4\tan \theta =21-2x$

$\Rightarrow 5\tan \theta =21-2x$

$\Rightarrow \tan \theta =\frac{21-2x}{5}$ ........$\left(3\right)$

$2\times$eqn$\left(1\right)-3\times$eqn$\left(2\right)$ we get

$\Rightarrow 4\sec \theta +6\tan \theta -9\sec \theta -6\tan \theta =14-3x$

$\Rightarrow -5\sec \theta =14-3x$

$\Rightarrow \sec \theta =\frac{14-3x}{-5}$ .......$\left(4\right)$

We know that ${\sec }^2\theta -{\tan }^2\theta =1$ .......$\left(5\right)$

Substituting $\left(3\right)$ and $\left(4\right)$ in $\left(5\right)$ we get

${\left(\frac{14-3x}{-5}\right)}^2-{\left(\frac{21-2x}{5}\right)}^2=1$

$\Rightarrow {(14-3x)}^2+{(21-2x)}^2=25$

$\Rightarrow (14-3x-21+2x)(14-3x+21-2x)=25$

$\Rightarrow (-x-7)(-5x+35)=25$

$\Rightarrow -(x+7)(-5)(x-7)=25$

$\Rightarrow x^2-49=5$

$\Rightarrow x^2=5+49=54$

$\therefore x=3\sqrt{6}$