If: `sin^(-1)x = alpha + beta` and `sin^(-1)y = alpha - beta`, then: `1+xy`=

Options

A. `sin^(2)alpha + sin^(2)beta`
B. `sin^(2)alpha + cos^(2)beta`
C. `cos^(2)alpha + cos^(2)beta`
D. `cos^(2)alpha + sin^(2)beta`

Answer

To solve the problem, we start with the given equations:

1. \( \sin^{-1} x = \alpha + \beta \)
2. \( \sin^{-1} y = \alpha - \beta \)

From these equations, we can express \( x \) and \( y \) in terms of sine:

\[ x = \sin(\alpha + \beta) \]
\[ y = \sin(\alpha - \beta) \]

Next, we need to find \( 1 + xy \):

\[ 1 + xy = 1 + \sin(\alpha + \beta) \sin(\alpha - \beta) \]

Using the identity for the product of sines, we have:

\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]

Applying this identity, we get:

\[ \sin(\alpha + \beta) \sin(\alpha - \beta) = \frac{1}{2} [\cos((\alpha + \beta) - (\alpha - \beta)) - \cos((\alpha + \beta) + (\alpha - \beta))] \]

This simplifies to:

\[ \sin(\alpha + \beta) \sin(\alpha - \beta) = \frac{1}{2} [\cos(2\beta) - \cos(2\alpha)] \]

Thus, we can rewrite \( 1 + xy \):

\[ 1 + xy = 1 + \frac{1}{2} [\cos(2\beta) - \cos(2\alpha)] \]

Now, simplifying this expression:

\[ 1 + xy = 1 + \frac{1}{2} \cos(2\beta) - \frac{1}{2} \cos(2\alpha) \]

To combine these terms, we can express 1 as \( \frac{2}{2} \):

\[ 1 + xy = \frac{2}{2} + \frac{1}{2} \cos(2\beta) - \frac{1}{2} \cos(2\alpha) \]

This leads us to:

\[ 1 + xy = \frac{2 + \cos(2\beta) - \cos(2\alpha)}{2} \]

We can also express \( \sin^2 \alpha + \cos^2 \beta \) using the Pythagorean identity:

\[ \sin^2 \alpha + \cos^2 \beta = 1 \]

Thus, we conclude that:

\[ 1 + xy = \sin^2 \alpha + \cos^2 \beta \]

**Final Answer:**
\[ 1 + xy = \sin^2 \alpha + \cos^2 \beta \]

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