Question

tan²(1/2sin⁻¹ 2/3), Evaluate it.

Answer 1

we have to find the value of tan²{1/2sin^-1(2/3)}

Let 1/2sin^-1(2/3) = A
sin^-1(2/3) = 2A
sin2A = 2/3

we know, $sin2\alpha=\frac{2tan\alpha}{1+tan^2\alpha}$

so, sin2A = 2/3 = 2tanA/(1 + tan²A)

1/3 = tanA/(1 + tan²A)

3tanA = 1 + tan²A

tan²A - 3tanA + 1 = 0

tanA = {3 ± √(9 - 4)}/2 [ using quadratic formula]

tanA = (3 ± √5)/2

we see sin^-1(2/3) belongs to [π/2, π/6]

so, (1/2sin^-1(2/3)) belongs to [π/12, π/4]

but in π/12 ≤ (1/2sin^-1(2/3) ≤ π/4 , tan will be less then 1

hence, tanA ≠ (3 + √5)/2

so, tanA = (3 - √5)/2 .........(1)

now, tan²{1/2sin^-1(2/3)} = tan²A

from equation (1),

tan²{1/2sin^-1(2/3)} = tan²A = (3- √5)²/4