જો frac{2sin alpha}{1 + cos alpha + sin alpha | vedclass

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(B) આપેલ છે,$\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$. અડધા ખૂણાના નિત્યસમનો ઉપયોગ કરતા: $\frac{2(2\sin \frac{\alpha}{2} \cos \frac{\al

Vedclass · Accepted Answer

$y$

Vedclass · Answer

(B) આપેલ છે,$\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$. અડધા ખૂણાના નિત્યસમનો ઉપયોગ કરતા: $\frac{2(2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2})}{2\cos^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = y$ $\frac{2\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2}} = y$. હવે,પદાવલિ $E = \frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha}$ લો. નિત્યસમનો ઉપયોગ કરતા,આપણને મળે છે: $E = \frac{2\sin^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2} = \frac{2\sin \frac{\alpha}{2}(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})}{(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2} = \frac{2\sin \frac{\alpha}{2}}{\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}} = y$.

જો $\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$ હોય,તો $\frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha} = $

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