Question

if cosec A =7/4,prove that 1+tan²A=sec²A

Answer 1

Hii ☺ !!

Cosec A = 7 /4 = Hypotenuse / Perpendicular

Hypotenuse = 7 and Perpendicular = 4

Therefore,

Base = √( Hypotenuse)² - (perpendicular)²)

Base = √7² - 4²

Base = √33

Therefore,

TanA = P/b = 4/√33 and sec A = h/b = 7/√33

Therefore,

1 + Tan²A = 1 + ( 4/√33)² = 1 + 16 / 33 = ( 33 + 16 /33) = 49/33.


And,

sec²A = ( 7/√33)² = 49 / 33.



Hence,

1 + Tan²A = sec²A.

Answer 2

HELLO DEAR,



GIVEN:- cosecA = 7/4
it means, sinA = 1/cosecA = 4/7

cosA = √(1 - sin²A) = √(1 - 16/49) = √33/7 = 1/secA

secA = 7/√33

so, tanA = sinA/cosA = (4/7)/(√33/7) = 4/√33


now, 1 + tan²A = 1 + (4/√33)²

=> 1 + 16/33

=> (33 + 16)/33

=> 49/33

AND, sec²A = (7/√33)² = 49/33

hence, L.H.S = R.H.S


I HOPE IT'S HELP YOU DEAR,
THANKS