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Question

Show that $$\sqrt{\displaystyle\dfrac{1-\cos A}{1+\cos A}}=\displaystyle\dfrac{\sin A}{1+\cos A}$$.


Solution

L.H.S. $$=\sqrt{\displaystyle\frac{1-\cos A}{1+\cos A}}$$
Multiplying by $$\sqrt{1+\cos A}$$ in numerator and denominator
$$=\sqrt{\displaystyle\frac{1-\cos A}{1+\cos A}}\times \sqrt{\displaystyle\frac{1+\cos A}{1+\cos A}}$$
$$=\sqrt{\displaystyle\frac{(1-\cos A)(1+\cos A)}{(1+\cos A)(1+\cos A)}}$$
$$=\sqrt{\displaystyle\frac{1-\cos^2A}{(1+\cos A)^2}}=\sqrt{\displaystyle\frac{\sin^2}{(1+\cos A)^2}}$$
$$=\displaystyle\frac{\sin A}{1+\cos A}=$$ R.H.S.

Mathematics

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