$sin A = \frac{4}{5}$ , $A$ being an acute angle. Find the value of $2 tan A + 3 sec A + 4 sec A \cdot csc A$ .

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Solution

Given $sin A = \frac{O p p}{H y p} = \frac{4}{5}$
In $△ A B C$ , ${A C}^{2} = {A B}^{2} + {B C}^{2}$ (By Pythagoras theorem)
$\Rightarrow {A B}^{2} = {A C}^{2} - {B C}^{2} = 5^{2} - 4^{2} = 25 - 16 = 9$
$∴ A B = \sqrt{9} = 3$
So $tan A = \frac{O p p}{A d j} = \frac{4}{3}$ , $sec A = \frac{H y p}{A d j} = \frac{5}{3}$ , $csc A = \frac{H y p}{O p p} = \frac{5}{4}$
By substituting the values in, $2 tan A + 3 sec A + 4 sec A \cdot csc A$
$= 2 (\frac{4}{3}) + 3 (\frac{5}{3}) + 4 (\frac{5}{3}) (\frac{5}{4})$
$= \frac{8}{3} + 5 + \frac{25}{3} = \frac{8 + 15 + 25}{3} = \frac{48}{3} = 16$
$∴ 2 tan A + 3 sec A + 4 sec A \cdot csc A = 16$

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