if 4cot A=3 and Ais acute, then
the value of sinA-cosA by sinA-cosA
Answers
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
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Answer:
CotA = 3/4 = base/perpendicular
Consider ∆ABC in which B is a right angled triangle and BA = 3 , CB = 4 and by Pythagoras theorem CA = 5
Now, we can find the value of sinA and cosA