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Question: If \(\alpha + \beta - \gamma = \pi,\) then\(\sin^{2}\alpha + \sin^{2}\beta - \sin^{2}\gamma =\)...

If α+β−γ=π,\alpha + \beta - \gamma = \pi, thensin⁡2α+sin⁡2β−sin⁡2γ=\sin^{2}\alpha + \sin^{2}\beta - \sin^{2}\gamma =

A

2sin⁡αsin⁡βcos⁡γ2\sin\alpha\sin\beta\cos\gamma

B

2cos⁡αcos⁡βcos⁡γ2\cos\alpha\cos\beta\cos\gamma

C

2sin⁡αsin⁡βsin⁡γ2\sin\alpha\sin\beta\sin\gamma

D

None of these

Answer

2sin⁡αsin⁡βcos⁡γ2\sin\alpha\sin\beta\cos\gamma

Explanation

Solution

We have α+β−γ=π.\alpha + \beta - \gamma = \pi.

Now sin⁡2α+sin⁡2β−sin⁡2γ\sin^{2}\alpha + \sin^{2}\beta - \sin^{2}\gamma =sin⁡2α+sin⁡(β−γ)sin⁡(β+γ)= \sin^{2}\alpha + \sin(\beta - \gamma)\sin(\beta + \gamma)

=sin⁡2α+sin⁡(π−α)sin⁡(β+γ)= \sin^{2}\alpha + \sin(\pi - \alpha)\sin(\beta + \gamma) (∵α+β−γ=π)(\because\alpha + \beta - \gamma = \pi)

=sin⁡2α+sin⁡αsin⁡(β+γ)=sin⁡α{sin⁡α+sin⁡(β+γ)}= \sin^{2}\alpha + \sin\alpha\sin(\beta + \gamma) = \sin\alpha\{\sin\alpha + \sin(\beta + \gamma)\}

=sin⁡α{sin⁡(π−β+γ)‾+sin⁡(β+γ)}= \sin\alpha\{\sin(\pi - \overline{\beta + \gamma)} + \sin(\beta + \gamma)\}

=sin⁡α{−sin⁡(γ−β)+sin⁡(γ+β)}= \sin\alpha\{ - \sin(\gamma - \beta) + \sin(\gamma + \beta)\}

=sin⁡α{2sin⁡βcos⁡γ}=2sin⁡αsin⁡βcos⁡γ= \sin\alpha\{ 2\sin\beta\cos\gamma\} = 2\sin\alpha\sin\beta\cos\gamma.