Question

Prove that 1+secA/secA = sin²A/(1-cosA)

Answer 1

Answer:

Step-by-step explanation:

To prove :  

secA

1+secA

​  

=  

1−cosA

sec  

12

A

​  

 

LHS  

secA

1+secA

​  

=  

cosA

1

​  

 

1+  

cosA

1

​  

 

​  

 

=  

cosA

1

​  

 

cosA

cosA+1

​  

 

​  

 

=  

cosA

cosA+1

​  

×  

1

cosA

​  

 

=1+cosA

=1+cosA×  

1−cosA

1−cosA

​  

 

=  

1−cosA

(1)  

2

−(cosA)  

2

 

​  

 

=  

1−cosA

1−cos  

2

A

​  

 

=  

1−cosA

sin  

2

A

​  

 

=RHS

To prove :  

secA

1+secA

​  

=  

1−cosA

sec  

12

A

​  

 

LHS  

secA

1+secA

​  

=  

cosA

1

​  

 

1+  

cosA

1

​  

 

​  

 

=  

cosA

1

​  

 

cosA

cosA+1

​  

 

​  

 

=  

cosA

cosA+1

​  

×  

1

cosA

​  

 

=1+cosA

=1+cosA×  

1−cosA

1−cosA

​  

 

=  

1−cosA

(1)  

2

−(cosA)  

2

 

​  

 

=  

1−cosA

1−cos  

2

A

​  

 

=  

1−cosA

sin  

2

A

​  

 

=RHS

Answer 2

Step-by-step explanation:

To Prove:-

• $\rm \dfrac{1 + secA}{secA} = \dfrac{ {sin}^{2} a}{1 - cosA}$

Proof:-

$\rm LHS = \dfrac{1 + secA}{secA}$

$\rm \dfrac{1 + \dfrac{1}{cosA} }{ \dfrac{1}{cosA} }$

$\rm = \dfrac{1 + \dfrac{1}{cosA} }{ \dfrac{1}{cosA} } \: \: \: \: \: \: \: \: \: \: \: \bigg( \because secA = \dfrac{1}{cosA} \bigg)$

$\rm = \dfrac{\dfrac{1 + cosA}{cosA} }{ \dfrac{1}{cosA} }$

$\rm = \dfrac{1 + cosA}{cosA} \times \dfrac{cosA}{1}$

$\rm = 1 + cosA$

By rationalising :-

$\rm = 1 + cosA \times \dfrac{1 - cosA}{1 - cosA}$

$\rm = \dfrac{(1 + cosA)(1 - cosA)}{1 - cosA}$

$\rm = \dfrac{ {(1)} ^{2} - (cosA)^{2} }{1 - cosA} \: \: \: \: \: \: \: \: \: \bigg(\because (a+b)(a-b) = a^{2} - b^{2}\bigg)$

$\rm = \dfrac{ {1} - cos^{2} A}{1 - cosA}$

$\rm As, \: 1 - {cos}^{2} A= sin ^{2} A$

$\rm = \dfrac{ sin^{2} A}{1 - cosA}$

= RHS

LHS = RHS

Hence Proved