pls solve this question 2 tan^-1 (A)=
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Answered by
1
diff w r t x
d/dx(2tan^-1(A))
=0 bcoz derivative of constant is o
A is not variable so we cant diff w r t A.
A is constant
it will useful to u...plz mark
d/dx(2tan^-1(A))
=0 bcoz derivative of constant is o
A is not variable so we cant diff w r t A.
A is constant
it will useful to u...plz mark
Answered by
5
Hey friend, ☺
*1). 2 tan^-1 (A) = sin^-1 [2(A) / (1+A^2)] ,where |A|<=1.
*2). 2 tan^-1(A) = cos^-1 [ (1-A^2) / (1+A^2)], where A >=0.
*3). 2 tan^-1(A) = tan^-1 [ (2A/1-A^2)], where -1 <A<1.
Proof:
(1).
Let tan^A = y. Now,
sin^-1 [ (2A/ 1+A^2)]
= sin^-1 [ 2 tany/(1+tan^2y) ]
=sin^-1 (sin2y)
= 2y
= 2 tan^-1(A).
(2).
Also, cos^-1 [(1-A^2) / (1+A^2)]
= cos ^-1 [ (1-tan^2y) / (1+tan^2y)]
=cos^-1 (cos2y)
= 2y
=2tan^-1 (A)
Similarly 3rd can be worked out.
☺ HOPE IT HELPS YOU!!!!! ☺
*1). 2 tan^-1 (A) = sin^-1 [2(A) / (1+A^2)] ,where |A|<=1.
*2). 2 tan^-1(A) = cos^-1 [ (1-A^2) / (1+A^2)], where A >=0.
*3). 2 tan^-1(A) = tan^-1 [ (2A/1-A^2)], where -1 <A<1.
Proof:
(1).
Let tan^A = y. Now,
sin^-1 [ (2A/ 1+A^2)]
= sin^-1 [ 2 tany/(1+tan^2y) ]
=sin^-1 (sin2y)
= 2y
= 2 tan^-1(A).
(2).
Also, cos^-1 [(1-A^2) / (1+A^2)]
= cos ^-1 [ (1-tan^2y) / (1+tan^2y)]
=cos^-1 (cos2y)
= 2y
=2tan^-1 (A)
Similarly 3rd can be worked out.
☺ HOPE IT HELPS YOU!!!!! ☺
Ruhanika105:
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