Question

please answer this......​

Answer 1

Answer:

given

=> 3 sinA = 4 cos A

dividing by cos A on both side of the equation

=> 3 sinA/ cosA = 4cosA/ cosA

=> 3 tanA = 4

=> tanA = 4/3

then in right angle triangle

height= 4 and base = 3

then find hypotenuse

use Pythagoras theorem

$= > {hypotenuse}^{2} = {4}^{2} + {3}^{2}$

$= > {hypotenuse}^{2} = 16 + 9$

$= > {hypotenuse}^{2} = 25$

$= > hypotenuse = \sqrt{25}$

=> hypotenuse = 5

(1). sinA= height/ hypotenuse

=> sinA= 4/5

(2). cosA = base/ hypotenuse

=> cosA = 3/5

(3). secA = 1/ cosA

=> secA = 5/3

$=>{tan}^{2} a-{sec}^{2} a \\ = {( \frac{4}{3})}^{2}-{(\frac{5}{3}) }^{2}$

$= \frac{16}{9} - \frac{25}{9}$

$= \frac{16 - 25}{9}$

$= - \frac{9}{9} = - 1$

$= > {tan}^{2}a - {sec}^{2}a = - 1$