sir miscellaneous exercise..ques no 14.sry i could not type that
Attachments:
kvnmurty:
the question and answer are there itself. what is required?
Answers
Answered by
1
To solve tan^-1 [ (1-x)/(1+x) ] = 1/2 * tan^-1 x , x > 0
This is easily solved using the following formula. That is done in the solution there, i guess. That is a simple solution. You can in a little longer way without using that too.
tan (A - B) = (tan A - tan B) / (1 + tan A tan B)
So A - B = tan^-1 [ (tan A - tan B) / (1 + tan A tan B) ]
comparing with LHS, we find tan A = 1, so A = π/4 ; tan B = x So B = tan^-1 x
So A - B = π/4 - tan^-1 x = RHS = 1/2 tan^-1 x
3/2 tan^-1 x = π/4 tan^-1 x = 2π/3*4 = π/6
x = tan π/6 = 1/√3
Easy way to get the solution.
======================================================
Let L H S = R H S = Ф
As L H S = Ф, tan Ф = (1-x) / (1- x)
This is easily solved using the following formula. That is done in the solution there, i guess. That is a simple solution. You can in a little longer way without using that too.
tan (A - B) = (tan A - tan B) / (1 + tan A tan B)
So A - B = tan^-1 [ (tan A - tan B) / (1 + tan A tan B) ]
comparing with LHS, we find tan A = 1, so A = π/4 ; tan B = x So B = tan^-1 x
So A - B = π/4 - tan^-1 x = RHS = 1/2 tan^-1 x
3/2 tan^-1 x = π/4 tan^-1 x = 2π/3*4 = π/6
x = tan π/6 = 1/√3
Easy way to get the solution.
======================================================
Let L H S = R H S = Ф
As L H S = Ф, tan Ф = (1-x) / (1- x)
Similar questions