How to solve this theorem? 2 tan'x=sin'(2x/1+x sq.) here, '=inverse sq=square?

Since tan is a surjective function from (-pi/2, pi/2) to the real numbers, there is a y in the range (-pi/2, pi/2) such that x = tan y.

Left hand side = 2 tan' (tan y) = 2y (this works only because we chose y in the interval (-pi/2, pi/2), which is in the range of the tan function)

Right hand side

= sin' (2 tan y/(1 + tan^2 y))

= sin' (2 tan y / (sec ^ y))

= sin' (2 (sin y / cos y) * cos^2 y)

= sin' (2 sin y cos y)

= sin' (sin 2y)

= 2y

The last step works because 2y is in the range (-pi, pi), which is in the range of sin'.

Since both sides are 2y, they are equal.

let 2 tan^-1(x) = A -----------------------(1)

tan^-1(x) = A/2

tan(A/2) = x

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

= 2x / (1 - x^2)

=> opposite side / adjacent side = 2x /(1 - x^2)

opposite side = 2x

adj.side = 1 - x^2

hypotenuse = √[(2x)^2 + (1 - x^2)^2 ]

= √4x^2 + 1 + x^4 - 2x^2 ]

=√[x^4 + 2x^2 + 1 ]

= (1 + x^2)

sin A = opposite side / hypotenuse

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

A = sin^-1 [2x /(1 + x^2) ] -------------------(2)

from (1) and (2)

2 tan^-1(x) = sin^-1 [2x /(1 + x^2) ]

a million. element tanx tanx (tanx - sqrt(3) = 0 now set each and each element to 0 so a) tanx = 0 or b) tanx - sqrt(3) = x= 0 tanx=sqrt(3) do inverse of tan x = 60 4. element cosx cosx (a million-2sinx) = 0 so a) cosx = 0 or b) a million - 2sinx=0 x-ninety a million=2sinx sinx=a million/2 x=30

Let,

2 tan'x = sin'(2x/1+x sq.) = ϴ

SO,

2x/(1+x^2) = sinϴ ........................................ [1]

and

x = tan(ϴ/2) .............................................. [2]

From [1] and [2], we have,

2tan(ϴ/2)/[1 + (tan(ϴ/2))^2 = sinϴ

tan(ϴ/2 +ϴ/2) = tanϴ = sinϴ

sinϴ/cosϴ = sinϴ

sinϴ -sinϴcosϴ = 0

sinϴ(1-cosϴ) = 0

sinϴ = 0, ====> ϴ = 0

1-cosϴ = 0, ===> cosϴ=1,

ϴ = 0,

Hence,

From [1], 2x/(1+x^2) = 0,

2x=0, ===> x= 0 >==============< ANSWER