Prove the identity $$\dfrac{\sin (2α -β) }{\sinα } -2\cos(α -β )=-\sinβ\cscα $$

Question

Prove the identity $$\dfrac{\sin (2α -β) }{\sinα  } -2\cos(α -β )=-\sinβ\cscα  $$

Uniteasy Admin · Accepted Answer

$$\dfrac{\sin (2α -β) }{\sinα } -2\cos(α -β )=-\sinβ\cscα $$
$$\dfrac{\sin (2α -β) }{\sinα } -2\cos(α -β )=- \dfrac{\sinβ}{\sinα}  $$
$$\sin(2α-β)-2\sinα\cos(α-β)=-\sinβ $$n
$$\sin(2α-β)-2\sinα\cos(α-β)$$
$$=\sinα\cos(α-β)+\cosα\sin(α-β)-2\sinα\cos(α-β) $$
$$=\cosα\sin(α-β)- \sinα\cos(α-β) $$
$$=\sin(α-β -α) =-\sinβ $$

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