Question

Sin5A/SinA- Cos5A/CosA=4Cos2A

Answer 1

Step-by-step explanation:

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Answer 2

To prove: $\dfrac{\sin 5A}{\sin A}- \dfrac{cos 5A}{\cos A} = 4 \cos 2A$

Step-by-step explanation:

Consider L.H.S.

We have

$\dfrac{\sin 5A}{\sin A}- \dfrac{\cos 5A}{\cos A} \\\\= \dfrac{\sin 5A \cos A-\cos 5A \sin A}{\sin A \cos A}$

Now as we know $\sin(x-y)= \sin x \cos y -\cos x \sin y$

Therefore we get

$=\dfrac{\sin (5A-A)}{\cos A \sin A} \\\\=\dfrac{\sin 4A}{\cos A \sin A} \\\\ \text {Now } \because \sin 2A = 2 \sin A \cos A \text { we get }\\\\ =\dfrac{2\sin 2A \cos 2A}{\cos A \sin A} \\\\= \dfrac{2 \times 2 \sin A \cos A\cos 2A}{\cos A \sin A} =4 \cos A$

which is equal to R.H.S.

Hence, proved the required result that  $\dfrac{\sin 5A}{\sin A}- \dfrac{cos 5A}{\cos A} = 4 \cos 2A$