Question

If $5\sin A = 3$ then the value of ${\sec ^2}A - {\tan ^2}A$ is:
A. $0$
B. $5$
C. $3$
D. $1$

Answer 1

Hint: Here we need to proceed by the knowledge of the formula of the trigonometric functions where in the right angled triangle we can say that
The value of $\sin A = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}$ and then we will get the value of the base by the Pythagoras theorem.
Now we can easily find all other trigonometric functions and get the desired result.

Complete step-by-step answer:
Here we are given that $5\sin A = 3$
So rearranging it we get that $\sin A = \dfrac{3}{5}$
We know that the formula of the trigonometric functions in the right angles triangle are as:

$\sin A = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}$
$\cos A = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}$
$\tan A = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$
$\cot A = \dfrac{{{\text{base}}}}{{{\text{perpendicular}}}}$
$\sec A = \dfrac{{{\text{hypotenuse}}}}{{{\text{base}}}}$
$\cos {\text{ec}}A = \dfrac{{{\text{hypotenuse}}}}{{{\text{perpendicular}}}}$
So we can accordingly put the values and get the desired result.
So we have got that $\sin A = \dfrac{3}{5}$
Also we know that $\sin A = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}$
So equating we can get that $\sin A = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}} = \dfrac{3}{5}$
So we can say that in the simplest form we have
Perpendicular$ = 3$
Hypotenuse$ = 5$
Now by Pythagoras theorem we get that
${\text{hypotenuse}}{{\text{e}}^2} = {\text{bas}}{{\text{e}}^2} + {\text{perpendicular}}{{\text{r}}^2}$
So now we can substitute the values and get:
$
\Rightarrow {5^2} = {\text{bas}}{{\text{e}}^2} + {3^2} \\
\Rightarrow {\text{bas}}{{\text{e}}^2} = {5^2} - {3^2} \\
\Rightarrow {\text{bas}}{{\text{e}}^2} = 25 - 9 = 16 \\
\Rightarrow {\text{base}} = \sqrt {16} = 4 \\
 $
Hence we have got that
Perpendicular$ = 3$
Hypotenuse$ = 5$
Base$ = 4$
So now we can easily calculate the value of $\tan A{\text{ and }}\sec A$ which are required.
$\Rightarrow$ $\sec A = \dfrac{{{\text{hypotenuse}}}}{{{\text{base}}}} = \dfrac{5}{4}$
$\Rightarrow$ $\tan A = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}} = \dfrac{3}{4}$
So we have got both the value which we need to solve the given problem:
Now we need to find the value that is asked which is ${\sec ^2}A - {\tan ^2}A$
Now we simply need to substitute their values which we have calculated and get our required result.
So substituting the values we get
$\Rightarrow$ ${\sec ^2}A$$ = {\left( {\dfrac{5}{4}} \right)^2} = \dfrac{{25}}{{16}}$
$\Rightarrow$ ${\tan ^2}A = {\left( {\dfrac{3}{4}} \right)^2} = \dfrac{9}{{16}}$
Now we can easily subtract above two equations and get the required value.
$\Rightarrow$ ${\sec ^2}A - {\tan ^2}A$$ = \dfrac{{25}}{{16}} - \dfrac{9}{{16}} = \dfrac{{25 - 9}}{{16}} = \dfrac{{16}}{{16}} = 1$
Hence we get that the value required of the trigonometric function ${\sec ^2}A - {\tan ^2}A$ is $1$
So D is the correct option.

Note: Here we must know what the formula of the trigonometric functions is in the right angled triangle like $\tan A = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$ and we must know how to utilise the Pythagoras theorem in order to calculate the values of hypotenuse, base and perpendicular in the right angled triangle.