Question

The solution set of equation ${{\sin }^{-1}}x=2{{\tan }^{-1}}x$ is
a)$\left\{ 1,2 \right\}$
b)$\left\{ -1,2 \right\}$
c)$\left\{ -1,1,0 \right\}$
d)$\left\{ 1,\dfrac{1}{2},0 \right\}$

Answer 1

Hint: When an equation is given in terms of sine, cosine, tangent, we must use any of the trigonometric identities to make the inequation solvable. There are inter relations between sine, cosine, tan, secant these are inter relations are called as identities. Whenever you can see conditions such that \[\theta \in R\] , that means inequality is true for all angles. So, directly think of identity which will make your work easy.

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] for all \[\theta \] , \[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]

Complete step-by-step answer:
An equality with sine, cosine or tangent in them is called a trigonometric equation. These are solved by some inter relations known beforehand.

All the inter relations which relate sine, cosine, tangent, secant, cotangent are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us nearer to result.

Given equation in the question which we need to find solution

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

Convert the tan term into the terms of sin; we get

By basic knowledge of inverse trigonometry, we can write

\[2{{\tan }^{-1}}x={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)\]

By substituting this equation in the main equation, we get:

\[{{\sin }^{-1}}x={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)\]

By applying sin on both sides of the equation, we get:

$x=\dfrac{2x}{1+{{x}^{2}}}$

By cross multiplying terms in the equation above, we get:

$x+{{x}^{5}}=2x$

By subtracting 2x on both sides of equation above, we get:

${{x}^{3}}-x=0$

By factoring the terms in above equation, we get that:

$x\left( x-1 \right)\left( x+1 \right)=0$

By using the above equation, we can say the roots of equation:

$x=0,1,-1$

So, the range of x is $\left\{ -1,1,0 \right\}$

Option (c) is the correct answer.

Note: Be careful while applying ${{\sin }^{-1}}$ as it may lead to extra roots. The conversion of everything into sin is done to cancel all angular terms and solve for x.