The solutions of the equation $sin^{-1}

CUET Preparation Today

Start Free Mocks

Home Questions 6T35G3PXG5

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

The solutions of the equation $sin^{-1} x = 2 tan^{-1}x.$

Options:

{1, 2}

{-1, 2}

{-1, 1, 0}

{1, 1/2, 0}

Correct Answer:

{-1, 1, 0}

Explanation:

Clearly , LHS of the given equation is meaningful for x ∈ [-1, 1] and RHS is defined for all x ∈ R. So, the given equation exists for x ∈ [-1,1].

Now,

$sin^{-1}x = 2 tan^{-1}x $

$⇒sin^{-1}x = sin^{-1} \left(\frac{2x}{1+x^2}\right)$

$⇒ x^3 = x ⇒ x =0 , 1, -1.$