Math, asked by Kushi4you, 6 months ago

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

Answers

Answered by shreya5btakshasila
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

Answered by MaIeficent
19

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

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