If sin alpha - cos alpha = m and sin 2 alpha | vedclass

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(B) Given that $\sin \alpha - \cos \alpha = m$ and $\sin 2 \alpha = n - m^2$.Squaring the first equation: $(\sin \alpha - \cos \alpha)^2 = m^2$.$\sin^

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$1$

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(B) Given that $\sin \alpha - \cos \alpha = m$ and $\sin 2 \alpha = n - m^2$.Squaring the first equation: $(\sin \alpha - \cos \alpha)^2 = m^2$.$\sin^2 \alpha + \cos^2 \alpha - 2 \sin \alpha \cos \alpha = m^2$.Since $\sin^2 \alpha + \cos^2 \alpha = 1$ and $2 \sin \alpha \cos \alpha = \sin 2 \alpha$,we have:$1 - \sin 2 \alpha = m^2$.$\sin 2 \alpha = 1 - m^2$.Comparing this with the given equation $\sin 2 \alpha = n - m^2$,we get:$n - m^2 = 1 - m^2$.Therefore,$n = 1$.

If $\sin \alpha - \cos \alpha = m$ and $\sin 2 \alpha = n - m^2$,where $-\sqrt{2} \leq m \leq \sqrt{2}$,then $n$ is equal to

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