Question

cosecA+cotA=3/2,then cosA

Answer 1

√5÷3 it is answer of your question

Answer 2

Hey buddy here is Solution :

Given

cot A + cosec A = 3

It can be written as

(cos A/sin A) + (1/sin A) = 3

(cos A + 1) / sin A = 3

cos A + 1 = 3 sin A

cos A = cos^2 (A/2) - sin^2 (A/2)

cos^2 (A/2) + sin^2 (A/2) = 1

cos^2 (A/2) - sin^2 (A/2) + cos^2 (A/2) + sin^2 (A/2) = 3*2 sin (A/2)cos (A/2),

2 cos^2 (A/2) = 3*2 sin (A/2)cos (A/2)

On solving

cos (A/2) = 3sin (A/2)

tan (A/2) = 1/3

tan A = 2tan (A/2)/[1 - tan^2 A]

● tan A = 2*(1/3)/(1 - 1/9)

● tan A = (2/3)/(8/9)

● tan A = (2*9)/ (3*8)

● tan A = 3/4

Using Pythagoras theorem here,

height of right-angled triangle is 3, its base is 4, hence the hypotenuse is 5

Therefore,

cos A = 4/5 and sin A = 3/5 .


hope solution will be helpful to u.....


plz mark my answer as brainalist one...