Question

If 3cotA = 4, examine whether 1-tan²A/1+tan²A =cos²A–sin²A.

Answer 1

Solve this as shown in the attachment.

Answer 2

Given, 3cotA = 4 => cotA = 4/3

tanA = 1/cotA = 3/4 = perpendicular/base

perpendicular = 3 , base = 4

from Pythagoras theorem,

hypotenuse = √{perpendicular² + base²}

= √{3² + 4²} = √{9 + 16} = 5

then, sinA = perpendicular/hypotenuse = 3/5

cosA = base/hypotenuse = 4/5

now, LHS = (1 - tan²A)/(1 + tan²A)

= [1 - (3/4)²]/[1 + (3/4)²]

= [1 - 9/16]/[1 + 9/16]

= 7/25

and RHS = cos²A - sin²A

= {4/5}² - {3/5}²

= 16/25 - 9/25

= 7/25

here it is clear that LHS = RHS , hence verified