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