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Question
If α + β − γ = π and sin2 α +sin2 β − sin2 γ = λ sin α sin β cos γ, then write the value of λ.
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Solution
\[\text{ Given }:\]
\[\gamma = - \left[ \pi - (\alpha + \beta) \right]\]
\[\text{ Also }, \]
\[\lambda = \frac{\sin^2 \alpha + \sin^2 \beta - \sin^2 \left[ - (\pi - (\alpha + \beta) \right]}{\sin\alpha \sin\beta \cos( - (\pi - (\alpha + \beta))} \]
\[ = \frac{\sin^2 \alpha + \sin^2 \beta - (\sin(\alpha + \beta) )^2}{- (\sin\alpha \sin\beta\cos(\alpha + \beta))} \left[ \sin \left( \pi - \theta \right) = \sin \theta and \cos\left( \pi - \theta \right) = - \cos \theta \right]\]
\[ = \frac{\sin^2 \alpha + \sin^2 \beta - \sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta - 2\sin\alpha \sin\beta \cos\alpha \cos\beta}{- (\sin\alpha \sin\beta \cos\alpha \cos\beta - \sin^2 \alpha \sin^2 \beta)}\]
\[ = \frac{\sin^2 \alpha(1 - \cos^2 \beta) + \sin^2 \beta(1 - \cos^2 \alpha) - 2\sin\alpha \sin\beta \cos\alpha \cos\beta}{\sin^2 \alpha \sin^2 \beta - \sin\alpha \sin\beta \cos\alpha \cos\beta}\]
\[ = \frac{2 \sin^2 \alpha \sin^2 \beta - 2\sin\alpha \sin\beta \cos\alpha \cos\beta}{\sin^2 \alpha \sin^2 \beta - \sin\alpha \sin\beta \cos\alpha \cos\beta}\]
\[ = 2\]
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