Question

Prove that:


$\sf \sqrt{\dfrac{1 - cos(\alpha)}{1 + cos(\alpha)}} = \dfrac{sin(\alpha)}{1 + cos(\alpha)}$

Answer 1

Step-by-step explanation:

Taking L.H.S,

$\sf \: \sqrt{ \dfrac{1 - \cos(\alpha) }{1 + \cos(\alpha) } } \\ \\ = \sf \sqrt{ \dfrac{\bigg(1 - \cos(\alpha)\bigg)\bigg(1 + \cos(\alpha) \bigg) }{\bigg(1 + \cos(\alpha)\bigg)\bigg(1 + \cos(\alpha) \bigg) } }$

Use the identities :

• (a+b)(a-b)=a²-b²
• (a+b)(a+b) = (a+b)²

$= \sf\sqrt{ \dfrac{1 - {cos}^{2} (\alpha)}{ {\bigg(1 + \cos(\alpha)\bigg) }^{2} } }$

$= \sf \: \sqrt{ \dfrac{ {sin}^{2} (\alpha)}{ {\bigg(1+ cos(\alpha)\bigg)}^{2} } } \\ \\ = \sf \dfrac{ \sin(\alpha) }{1 + \cos(\alpha) } \: \{proved \}$

_______________

Some formulas :-

★ sin²A + cos²A = 1

★ 1 + tan²A = sec²A

★1 + cot²A= cosec²A

★ cos²A - sin²A = cos2A

★ sin(A+B) = sinAcosB + cosAsinB

★ sin(A-B) = sinAcosB - cosAsinB

★ cos(A+B) = cosAsinB- sinAsinB

★ cos(A-B) = cosAsinB + sinAsinB

★ $\sf{tan(A+B)=\dfrac{tanA+tanB}{1-tanAtanB}}$

★$\sf{tan(A-B)=\dfrac{tanA-tanB}{1+tanAtanB}}$

★ sin2∅ = 2sin∅cos∅

★ cos2∅ = 2cos²∅ - 1

★ cos2∅ = 1 - 2sin²∅

Answer 2

Step-by-step explanation:

Taking L.H.S,

1 - cos(a)

V1+ cos(a)

1+ cos(a) 1+cos(a) 1+ cos(a)

1-cos(a)

Use the identities :

• (a+b)(a-b)=a²-b? (a+b)(a+b) = (a+b)

1 COS-a

1+cos(a)

sin (a)

2 1+ cos(a)

sin(a) 1+ cos(a) {proved}

Some formulas :

* sin2A + cos?A = 1

* 1 + tan?A = sec?A =

*1 + cotA= cosec2A

* cos?A - sin?A = cos2A

* sin(A+B) = sinAcosB +

cosAsinB

* sin(A-B) = sinAcosB -

cosAsinB

* cos(A+B) = cosAsinB sinAsinB

* cos(A-B) = cosAsinB + sinAsinB

* tan(A + B)

tanA + tanB 1- tanAtanB

*tan(A-B) = tanA - tanB

1+tanAtanB

* sin20 = 2sin Øcos

* cos20 = 2cos2 0-1

* cos20 = 1 - 2 sin20