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

If 5sin⁡A=35\sin A = 3 then the value of sec⁡2A−tan⁡2A{\sec ^2}A - {\tan ^2}A is:
A. 00
B. 55
C. 33
D. 11

Explanation

Solution

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=perpendicularhypotenuse\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 5sin⁡A=35\sin A = 3
So rearranging it we get that sin⁡A=35\sin A = \dfrac{3}{5}
We know that the formula of the trigonometric functions in the right angles triangle are as:

sin⁡A=perpendicularhypotenuse\sin A = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}
cos⁡A=basehypotenuse\cos A = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}
tan⁡A=perpendicularbase\tan A = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}
cot⁡A=baseperpendicular\cot A = \dfrac{{{\text{base}}}}{{{\text{perpendicular}}}}
sec⁡A=hypotenusebase\sec A = \dfrac{{{\text{hypotenuse}}}}{{{\text{base}}}}
cos⁡ecA=hypotenuseperpendicular\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=35\sin A = \dfrac{3}{5}
Also we know that sin⁡A=perpendicularhypotenuse\sin A = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}
So equating we can get that sin⁡A=perpendicularhypotenuse=35\sin A = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}} = \dfrac{3}{5}
So we can say that in the simplest form we have
Perpendicular=3 = 3
Hypotenuse=5 = 5
Now by Pythagoras theorem we get that
hypotenusee2=base2+perpendicularr2{\text{hypotenuse}}{{\text{e}}^2} = {\text{bas}}{{\text{e}}^2} + {\text{perpendicular}}{{\text{r}}^2}
So now we can substitute the values and get:
⇒52=base2+32 ⇒base2=52−32 ⇒base2=25−9=16 ⇒base=16=4  \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 = 3
Hypotenuse=5 = 5
Base=4 = 4
So now we can easily calculate the value of tan⁡A and sec⁡A\tan A{\text{ and }}\sec A which are required.
⇒\Rightarrow sec⁡A=hypotenusebase=54\sec A = \dfrac{{{\text{hypotenuse}}}}{{{\text{base}}}} = \dfrac{5}{4}
⇒\Rightarrow tan⁡A=perpendicularbase=34\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⁡2A−tan⁡2A{\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⁡2A{\sec ^2}A =(54)2=2516 = {\left( {\dfrac{5}{4}} \right)^2} = \dfrac{{25}}{{16}}
⇒\Rightarrow tan⁡2A=(34)2=916{\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⁡2A−tan⁡2A{\sec ^2}A - {\tan ^2}A =2516−916=25−916=1616=1 = \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⁡2A−tan⁡2A{\sec ^2}A - {\tan ^2}A is 11
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=perpendicularbase\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.