Question

if tan=3/4 then,1-cosA/1+cosA​

Answer 1

Answer:

If tan A= 3/4 , (1-cos A)/(1+cos A) =?

(1-cos A)/(1+cos A) = (1- cos A)×(1- cos A)/(1+ cos A).(1 - cos A).

= (1- cos A)^2/(1-cos^2 A).

= {(1- cos A)/sin A}^2. , Dividing in Nr and Dr by cos A.

= {(sec A - 1)/tan A}^2.

= {+/-√(1+tan^2 A) - 1}^2/ tan^2 A. , [putting tan A = 3/4.]

= {+/- √(1+9/16) -1}^2/(9/16).

= {+/- 5/4 - 1}^2×(16/9).

= (5/4 -1)^2×(16/9) , (-5/4. -1)^2×(16/9).

= (1/16)×(16/9) , (81/16)×(

= { 1/sin A - cos A/sin A }^2.

= ( cosec A - cot A)^2.

= {+/- √(1+cot^2 A) - cot A }^2

Putting cot A = 4/3 , [ as tan A =3/4. given ].

= { +/- √(1 +16/9) - 4/3}^2.

= { +/- 5/3 - 4/3}^2.

= (5/3- 4/3)^2. , (-5/3–4/3)^2

= (1/3)^2. , (- 3)^2.

= 1/9 , 9 , Answer.

Step-by-step explanation:

Hope you understood

Answer 2

Hope this given image helps u