NIMCET 2012 — Mathematics PYQ

Q: How do I solve NIMCET 2012 Mathematics questions?

Practice NIMCET 2012 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

If $\cos(\alpha + \beta) = \frac{4}{5}$ and $\sin(\alpha - \beta) = \frac{5}{13}$ , where $0 < \alpha, \beta < \frac{\pi}{4}$ , then $\tan(2\alpha)$ is equal to:

Choose the correct answer:

Explanation

Step 1: Find $\tan(\alpha + \beta)$ Given $\cos(\alpha + \beta) = \frac{4}{5}$ . Since $0 < \alpha, \beta < \frac{\pi}{4}$ , the sum $(\alpha + \beta)$ lies in the first or second quadrant, but based on the constraints, it's in the first. Using the identity $\sin^2\theta + \cos^2\theta = 1$ :

\sin(\alpha + \beta) = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}

Therefore,

\tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{3/5}{4/5} = \frac{3}{4}

Step 2: Find $\tan(\alpha - \beta)$ Given $\sin(\alpha - \beta) = \frac{5}{13}$ . Using the same identity to find $\cos(\alpha - \beta)$ :

\cos(\alpha - \beta) = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}

Therefore,

\tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} = \frac{5/13}{12/13} = \frac{5}{12}

Step 3: Use the compound angle formula for $\tan(2\alpha)$ We can write $2\alpha$ as $(\alpha + \beta) + (\alpha - \beta)$ . Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ :

\tan(2\alpha) = \tan[(\alpha + \beta) + (\alpha - \beta)]

\tan(2\alpha) = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta) \tan(\alpha - \beta)}

Step 4: Substitute the values

\tan(2\alpha) = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \left(\frac{3}{4} \cdot \frac{5}{12}\right)}

\tan(2\alpha) = \frac{\frac{9 + 5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{\frac{48 - 15}{48}}

\tan(2\alpha) = \frac{14}{12} \times \frac{48}{33}

\tan(2\alpha) = \frac{14 \times 4}{33} = \frac{56}{33}

Conclusion: The value of $\tan(2\alpha)$ is $\frac{56}{33}$ .

Correct Option: (a)

Choose the correct answer:

Explore More