Question

all trigonometry formulla of 10th class

Answer 1

let the height of a right angle triangle be h
let the base of a right angle triangle be b
let the hypotenuse of a a right angle triangle be hy

so sin Ф = h/hy     cosec Ф = hy/h
    cos Ф = b/hy     sec Ф = hy/b
   tan Ф = h/b        cot Ф = b/h

so cos²Ф + sin²Ф = 1 
    1 + tan²Ф = sec²Ф
    1 + cot²Ф = cot²Ф
cos(90 -  Ф) = sin Ф
tan( 90 - Ф) = cot Ф
cosec (90 - Ф) = sec Ф

Answer 2

sin (90° – A) = cos A
cos (90° – A) = sin A
tan (90° – A) = cot A
cot (90° – A) = tan A
sec (90° – A) = cosec A
cosec(90° – A) = sec A.

sin² A + cos² A = 1
sec² A – tan² A = 1 for 0° ≤ A < 90°
cosec² A = 1 + cot² A for 0° < A ≤ 90º.