Question

prove that
\frac{ \sin( \alpha ) }{1 - \cos( \alpha ) } + \frac{1 + \cos( \alpha ) }{ \sin( \alpha ) } = 2 cosec \alpha
1−cos(α)
sin(α)
​
+
sin(α)
1−cos(α)
​
=2cosecα
prove this pleass. ​

Answer 1

Answer:

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Answer 2

$Step-by-step explanation:</p><p></p><p>Taking L.H.S,</p><p></p><p>\begin{gathered} \sf \: \sqrt{ \dfrac{1 - \cos(\alpha) }{1 + \cos(\alpha) } } \\ \\ = \sf \sqrt{ \dfrac{\bigg(1 - \cos(\alpha)\bigg)\bigg(1 + \cos(\alpha) \bigg) }{\bigg(1 + \cos(\alpha)\bigg)\bigg(1 + \cos(\alpha) \bigg) } }\end{gathered}1+cos(α)1−cos(α)=(1+cos(α))(1+cos(α))(1−cos(α))(1+cos(α))</p><p></p><p>Use the identities :</p><p></p><p>(a+b)(a-b)=a²-b²</p><p></p><p>(a+b)(a+b) = (a+b)²</p><p></p><p>= \sf\sqrt{ \dfrac{1 - {cos}^{2} (\alpha)}{ {\bigg(1 + \cos(\alpha)\bigg) }^{2} } }=(1+cos(α))21−cos2(α)</p><p></p><p>\begin{gathered}= \sf \: \sqrt{ \dfrac{ {sin}^{2} (\alpha)}{ {\bigg(1+ cos(\alpha)\bigg)}^{2} } } \\ \\ = \sf \dfrac{ \sin(\alpha) }{1 + \cos(\alpha) } \: \{proved \}\end{gathered}=(1+cos(α))2sin2(α)=1+cos(α)sin(α){proved}</p><p></p><p>_______________</p><p></p><p>Some formulas :-</p><p></p><p>★ sin²A + cos²A = 1</p><p></p><p>★ 1 + tan²A = sec²A</p><p></p><p>★1 + cot²A= cosec²A</p><p></p><p>★ cos²A - sin²A = cos2A</p><p></p><p>★ sin(A+B) = sinAcosB + cosAsinB</p><p></p><p>★ sin(A-B) = sinAcosB - cosAsinB</p><p></p><p>★ cos(A+B) = cosAsinB- sinAsinB</p><p></p><p>★ cos(A-B) = cosAsinB + sinAsinB</p><p></p><p>★ \sf{tan(A+B)=\dfrac{tanA+tanB}{1-tanAtanB}}tan(A+B)=1−tanAtanBtanA+tanB</p><p></p><p>★\sf{tan(A-B)=\dfrac{tanA-tanB}{1+tanAtanB}}tan(A−B)=1+tanAtanBtanA−tanB</p><p></p><p>★ sin2∅ = 2sin∅cos∅</p><p></p><p>★ cos2∅ = 2cos²∅ - 1</p><p></p><p>★ cos2∅ = 1 - 2sin²∅</p><p></p><p>$