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Prove:
\(\dfrac{\cos A}{1\pm \sin A}=\tan ({45}^{{}^{\circ }}\pm \dfrac{A}{2})\)
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Prove:
\(\dfrac{\cos A}{1\pm \sin A}=\tan ({45}^{{}^{\circ }}\pm \dfrac{A}{2})\)

Answer

See answer below
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Solution

Taking L.H.S.
\(\dfrac{\cos A}{1\pm \sin A}\)
Using trignometric identities:\(cosA=\;{cos}^{2}\dfrac{A}{2}−{sin}^{2}\dfrac{A}{2},\;sinA=2sin\dfrac{A}{2}cos\dfrac{A}{2}\)and \({cos}^{2}\dfrac{A}{2}+{sin}^{2}\dfrac{A}{2}=1\)
\(=\dfrac{{cos}^{2}\dfrac{A}{2}-{sin}^{2}\dfrac{A}{2}}{{cos}^{2}\dfrac{A}{2}+{sin}^{2}\dfrac{A}{2}\pm 2sin\dfrac{A}{2}cos\dfrac{A}{2}}\)
\(=\dfrac{\left(cos\dfrac{A}{2}-sin\dfrac{A}{2}\right)\left(cos\dfrac{A}{2}+cos\dfrac{A}{2}\right)}{{\left. \left(cos\dfrac{A}{2}\pm sin\dfrac{A}{2}\right)\right. }^{2}}\)
\(=\dfrac{cos\dfrac{A}{2}\pm sin\dfrac{A}{2}}{cos\dfrac{A}{2}\mp sin\dfrac{A}{2}}\)
\(=\dfrac{1\pm tan\dfrac{A}{2}}{1\mp tan\dfrac{A}{2}}\)
\(=\dfrac{tan\dfrac{\mathit{\pi }}{4}\pm tan\dfrac{A}{2}}{1\mp tan\dfrac{\mathit{\pi }}{4}\ldotp tan\dfrac{A}{2}}\)
\(=tan\left. \left(\dfrac{\mathit{\pi }}{4}\pm \dfrac{A}{2}\right)\right.\)
\(=tan\left. \left({\displaystyle {45}^{{}^{\circ }}}\pm \dfrac{A}{2}\right)\right.\)
Hence proved
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