NIMCET 2010 — Mathematics PYQ

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Question

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

If $\sin^{-1} \frac{2a}{1+a^2} - \cos^{-1} \frac{1-b^2}{1+b^2} = \tan^{-1} \frac{2x}{1-x^2}$ then $x$ is equal to:

Choose the correct answer:

Explanation

Step 1: Substitute the identities into the given equation.

Substituting these into the original equation:

2\tan^{-1} a - 2\tan^{-1} b = 2\tan^{-1} x

Step 2: Simplify the equation by dividing by 2.

\tan^{-1} a - \tan^{-1} b = \tan^{-1} x

Step 3: Apply the subtraction formula for $\tan^{-1}$ .

Using the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$ :

\tan^{-1} \left( \frac{a-b}{1+ab} \right) = \tan^{-1} x

Step 4: Solve for $x$ .

Comparing both sides, we get:

x = \frac{a-b}{1+ab}

Correct Option:

(d) $\frac{a-b}{1+ab}$