Question

Prove that:
$\frac{sin \: a}{1 + cos \: a} + \frac{sin \: a}{1 - cos \: a} = 2cosec \: a$
​

Answer 1

Solution

To Prove :-

• (sin a)/(1 + cos a) + (sin a )/(1 - cos a) = 2 cosec a

Prove

Take first L.H.S.

= (sin a)/(1 + cos a) + (sin a )/(1 - cos a)

Here ( 1 - cos² a) be LCM of these fraction

= [(sin a )( 1 - cos a) + (sin a )( 1 + cos a)]/(1 - cos² a)

We Know,

★( sin ² a + cos² a = 1)

= (sin a - sin a . cos a + sin a + sin a . cos a )/sin ² a

= 2 sin a / sin² a

= 2/sin a

We Know,

★ 1/sin a = cosec a

= 2 cosec a

That's proved

___________________

= R.H.S.

Answer 2

L.H.S:-

:$\implies$$\sf\dfrac{SinA}{1+CosA}$ + $\sf\dfrac{SinA}{1-CosA}$

:$\implies$$\sf\dfrac{SinA(1-CosA)+SinA(1+CosA}{(1+CosA)(1-CosA)}$

:$\implies$$\sf\dfrac{SinA-SinA.CosA+SinA+SinA.CosA}{1-Cos²A}$

:$\implies$$\sf\dfrac{2SinA}{Sin²A}$ ... {1-Cos²A = Sin²A}

:$\implies$$\sf\dfrac{2}{SinA}$ = $\sf 2CosecA$ = R.H.S

Hence, proved !!