Question

prove that sin A/1-cosA=cosec A+cotA
​

Answer 1

Step-by-step explanation:

Solution :

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Given :

To prove that :

⇒ \frac{Sin A}{1 + Cos A} = Cosec A - Cot A

1+CosA

SinA

=CosecA−CotA

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Proof :

LHS = \frac{Sin A}{1 + Cos A}

1+CosA

SinA

By multiplying with 1 - Cos A both the sides,

We get,

⇒ \frac{Sin A}{1 + Cos A} ( \frac{1 - Cos A}{1 - Cos A} )

1+CosA

SinA

(

1−CosA

1−CosA

)

⇒ \frac{Sin A (1 - Cos A)}{(1 + Cos A)(1 - Cos A)}

(1+CosA)(1−CosA)

SinA(1−CosA)

The denominator is in the form,

⇒ (a + b)(a - b),.

Hence,

We can use this identity : (a - b)(a + b) = a² - b²

⇒ \frac{Sin A (1 - Cos A)}{1^2 - Cos^{2} A }

1

2

−Cos

2

A

SinA(1−CosA)

⇒ \frac{Sin A (1 - Cos A)}{1 - Cos^{2} A}

1−Cos

2

A

SinA(1−CosA)

We know that,

⇒ Sin² A + Cos² A = 1

∴ Sin²A = 1 - Cos² A

⇒ \frac{Sin A (1 - Cos A)}{ Sin^2 A }

Sin

2

A

SinA(1−CosA)

⇒ \frac{Sin A (1 - Cos A)}{(Sin A)(Sin A)}

(SinA)(SinA)

SinA(1−CosA)

⇒ \frac{1 - Cos A}{Sin A}

SinA

1−CosA

⇒ \frac{1}{Sin A} - \frac{Cos A}{Sin A}

SinA

1

−

SinA

CosA

⇒ Cosec A - Cot ACosecA−CotA

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Hope it Helps !!

Answer 2

Answer:

L.H.S.= sin A/(1-cos A) . Multiply and divide with (1+cos A)

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

= sin A(1+cos A)/(1-cos^2 A)

= sin A(1+cos A)/sin^2 A

=(1+cos A)/sin A

= 1/sinA + cosA/sinA

= cosecA+cotA= RHS