Answer :

We are given that,


We need to find the value of x2.


Take,



Multiply on both sides by tangent.



Since, we know that tan(tan-1 x) = x.


So,



Or



Now, we need to simplify it in order to find x2. So, rationalize the denominator by multiplying and dividing by .




Note the denominator is in the form: (x + y)(x – y), where


(x + y)(x – y) = x2 – y2


So,


…(i)


Numerator:


Applying the algebraic identity in the numerator, (x – y)2 = x2 + y2 – 2xy.


We can write as,




Again using the identity, (x + y)(x – y) = x2 – y2.



…(ii)


Denominator:


Solving the denominator, we get




…(iii)


Substituting values of Numerator and Denominator from (ii) and (iii) in equation (i),





By cross-multiplication,


x2 tan α = 1 – √(1 – x4)


√(1 – x4) = 1 – x2 tan α


Squaring on both sides,


[√(1 – x4)]2 = [1 – x2 tan α]2


1 – x4 = (1)2 + (x2 tan α)2 – 2x2 tan α [, (x – y)2 = x2 + y2 – 2xy]


1 – x4 = 1 + x4 tan2 α – 2x2 tan α


x4 tan2 α – 2x2 tan α + x4 + 1 – 1 = 0


x4 tan2 α – 2x2 tan α + x4 = 0


Rearranging,


x4 + x4 tan2 α – 2x2 tan α = 0


x4 (1 + tan2 α) – 2x2 tan α = 0


x4 (sec2 α) – 2x2 tan α = 0 [, sec2 x – tan2 x = 1 1 + tan2 x = sec2 x]


Taking x2 common from both terms,


x2 (x2 sec2 α – 2 tan α) = 0


x2 = 0 or (x2 sec2 α – 2 tan α) = 0


But x2 ≠ 0 as according to the question, we need to find some value of x2.


x2 sec2 α – 2 tan α = 0


x2 sec2 α = 2 tan α


In order to find the value of x2, shift sec2 α to Right Hand Side (RHS).



Putting ,




x2 = 2 sin α cos α


Using the trigonometric identity, 2 sin x cos x = sin 2x.


x2 = sin 2α

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