If $3{sec}^2\theta +tan\theta -7=0$, 0° < θ < 90°, then what is the value of $\frac{2sin\theta +3cos\theta }{sec\theta +cosec\theta }$?

The correct answer is $\frac{5}{4}$.

Let's start with the given equation:

$3{sec}^2\theta +tan\theta -7=0$

We know the trigonometric identity ${sec}^2\theta =1+{tan}^2\theta$. We can substitute this into the given equation to make it an equation entirely in terms of $tan\theta$:

$3(1+{tan}^2\theta )+tan\theta -7=0$

Expanding the terms:

$3+3{tan}^2\theta +tan\theta -7=0$

Combining the constants:

$3{tan}^2\theta +tan\theta -4=0$

This is a quadratic equation where the variable is $tan\theta$. We can factor it by splitting the middle term. We need two numbers that multiply to $3\times (-4)=-12$ and add up to $1$. These numbers are $4$ and $-3$:

$3{tan}^2\theta +4tan\theta -3tan\theta -4=0$

Taking common factors from pairs of terms:

$tan\theta (3tan\theta +4)-1(3tan\theta +4)=0$

$(tan\theta -1)(3tan\theta +4)=0$

This gives us two possible solutions for $tan\theta$:

$tan\theta =1$ or $tan\theta =-\frac{4}{3}$

The problem states that $\theta$ is an acute angle, specifically $0°<\theta <90°$. In this first quadrant, all trigonometric functions, including $tan\theta$, are strictly positive. Therefore, we can discard the negative value:

$tan\theta =1$

The angle for which the tangent is $1$ is $45��$. So, $\theta =45°$.

Now, we are asked to find the value of the following expression:

$\frac{2sin\theta +3cos\theta }{sec\theta +cosec\theta }$

Substitute $\theta =45°$ into the expression:

$\frac{2sin45°+3cos45°}{sec45°+cosec45°}$

Using the standard values of trigonometric functions at $45°$:

$sin45°=\frac{1}{\sqrt{2}}$
$cos45°=\frac{1}{\sqrt{2}}$
$sec45°=\sqrt{2}$
$cosec45°=\sqrt{2}$

Plugging these values in, we get:

$\frac{2\left(\frac{1}{\sqrt{2}}\right)+3\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}+\sqrt{2}}$

Simplify the numerator and the denominator:

$\frac{\frac{5}{\sqrt{2}}}{2\sqrt{2}}$

This division can be rewritten as multiplication:

$\frac{5}{\sqrt{2}}\times \frac{1}{2\sqrt{2}}$

Multiplying the terms across:

$\frac{5}{2\times (\sqrt{2}\times \sqrt{2})}$

$=\frac{5}{2\times 2}$

$=\frac{5}{4}$