NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

If $\tan \theta = \frac{b}{a}$ , then the value of $a \cos 2\theta + b \sin 2\theta$ is:

Choose the correct answer:

Explanation

1. Recall the formulae:

$\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$
$\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$

2. Substitute $\tan \theta = \frac{b}{a}$ into the expression:

a \cos 2\theta + b \sin 2\theta

a \left( \frac{1 - (\frac{b}{a})^2}{1 + (\frac{b}{a})^2} \right) + b \left( \frac{2(\frac{b}{a})}{1 + (\frac{b}{a})^2} \right)

3. Simplify the terms inside the parentheses:

Denominator for both terms: $1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2}$

Numerator for first term: $1 - \frac{b^2}{a^2} = \frac{a^2 - b^2}{a^2}$

Numerator for second term: $\frac{2b}{a}$

Substituting these back:

a \left( \frac{\frac{a^2 - b^2}{a^2}}{\frac{a^2 + b^2}{a^2}} \right) + b \left( \frac{\frac{2b}{a}}{\frac{a^2 + b^2}{a^2}} \right)

a \left( \frac{a^2 - b^2}{a^2 + b^2} \right) + b \left( \frac{2b}{a} \times \frac{a^2}{a^2 + b^2} \right)

a \left( \frac{a^2 - b^2}{a^2 + b^2} \right) + b \left( \frac{2ab}{a^2 + b^2} \right)

4. Combine the expressions:

\frac{a(a^2 - b^2) + 2ab^2}{a^2 + b^2}

\frac{a^3 - ab^2 + 2ab^2}{a^2 + b^2}

\frac{a^3 + ab^2}{a^2 + b^2}

5. Factor out $a$ from the numerator:

\frac{a(a^2 + b^2)}{a^2 + b^2}

a

Conclusion:

The value is $a$ . The correct option is (b).

Choose the correct answer:

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