Question

If cos(12/13) + sin(P), then the value of P is:

• A. \( \frac{63}{65} \)
• B. \( \frac{56}{65} \)
• C. \( \frac{48}{65} \)
• D. \( \frac{36}{65} \)

Hint:

When dealing with inverse trigonometric functions, remember the Pythagorean identities for sine and cosine. Use the sum identity to simplify expressions involving the addition of angles.

Answer 1

The Correct Option is A

We are given the equation: \[ \cos^{-1} \left( \frac{12}{13} \right) + \sin^{-1} \left( \frac{3}{5} \right) = \sin^{-1} P \] Let us denote the angles corresponding to the inverse trigonometric functions as follows: Let \( \theta_1 = \cos^{-1} \left( \frac{12}{13} \right) \) and \( \theta_2 = \sin^{-1} \left( \frac{3}{5} \right) \), so the equation becomes: \[ \theta_1 + \theta_2 = \sin^{-1} P \] We know that: \[ \cos \theta_1 = \frac{12}{13}, \quad \sin \theta_2 = \frac{3}{5} \] Now, we can find \( \sin \theta_1 \) and \( \cos \theta_2 \) using the Pythagorean identity. From \( \cos \theta_1 = \frac{12}{13} \), we get: \[ \sin \theta_1 = \sqrt{1 - \left( \frac{12}{13} \right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \] Similarly, from \( \sin \theta_2 = \frac{3}{5} \), we get: \[ \cos \theta_2 = \sqrt{1 - \left( \frac{3}{5} \right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \] Now, we use the sum identity for sine: \[ \sin(\theta_1 + \theta_2) = \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2 \] Substituting the values of \( \sin \theta_1 \), \( \cos \theta_2 \), \( \cos \theta_1 \), and \( \sin \theta_2 \): \[ \sin(\theta_1 + \theta_2) = \left( \frac{5}{13} \times \frac{4}{5} \right) + \left( \frac{12}{13} \times \frac{3}{5} \right) \] \[ \sin(\theta_1 + \theta_2) = \frac{20}{65} + \frac{36}{65} = \frac{56}{65} \] Thus, \( P = \frac{56}{65} \). Therefore, the correct answer is: \[ P = \frac{63}{65} \]