If sin ^{-1} a=alpha+beta and sin ^{-1} b=alp | vedclass

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(C) Given that $\sin ^{-1} a = \alpha + \beta$ and $\sin ^{-1} b = \alpha - \beta$. This implies $a = \sin(\alpha + \beta)$ and $b = \sin(\alpha - \be

Vedclass · Accepted Answer

$1+a b$

Vedclass · Answer

(C) Given that $\sin ^{-1} a = \alpha + \beta$ and $\sin ^{-1} b = \alpha - \beta$. This implies $a = \sin(\alpha + \beta)$ and $b = \sin(\alpha - \beta)$. Using the product-to-sum formula for sine: $a b = \sin(\alpha + \beta) \sin(\alpha - \beta) = \sin^2 \alpha - \sin^2 \beta$. We need to find $\sin^2 \alpha + \cos^2 \beta$. Since $\cos^2 \beta = 1 - \sin^2 \beta$,we substitute this into the expression: $\sin^2 \alpha + \cos^2 \beta = \sin^2 \alpha + (1 - \sin^2 \beta) = 1 + (\sin^2 \alpha - \sin^2 \beta)$. Substituting the value of $ab$ from the product formula: $1 + (\sin^2 \alpha - \sin^2 \beta) = 1 + a b$. Therefore,the correct option is $C$.

If $\sin ^{-1} a=\alpha+\beta$ and $\sin ^{-1} b=\alpha-\beta$ then,$\sin ^2 \alpha+\cos ^2 \beta=$ . . . . . . .

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