Question

Prove that
1 + cos 0/
1 - cos
= ( cosec 0 + coto) square ​

Answer 1

Answer:

Step-by-step explanation:

cotθ−cscθ+1

cotθ+cscθ−1

​  

 

=  

cotθ−cscθ+1

cotθ+cscθ−(csc  

2

θ−cot  

2

θ)

​  

 

=  

cotθ−cscθ+1

cotθ+cscθ−(cscθ−cotθ)(cscθ+cotθ)

​  

 

=  

cotθ−cscθ+1

(cotθ+cscθ)(1−cscθ+cotθ)

​  

 

=cotθ+cscθ     .........Result1

=  

sinθ

cosθ

​  

+  

sinθ

1

​  

 

=  

sinθ

1+cosθ

​  

     .........Result2

Hence from result1 and result2 we have

cotθ−cscθ+1

cotθ+cscθ−1

​  

=cotθ+cscθ=  

sinθ

1+cosθ

​  

 

Hence proved.

(ii)L.H.S=  

secA−tanA

1

​  

−  

cosA

1

​  

 

Multiplying by secA+tanA in the numerator and denominator of first term,we get

=  

(secA−tanA)(secA+tanA)

secA+tanA

​  

−secA

=  

(sec  

2

A−tan  

2

A)

secA+tanA

​  

−secA

=secA+tanA−secA since sec  

2

A−tan  

2

A=1

=tanA

Adding and subtracting secA, we get

=secA+tanA−secA

=  

cosA

1

​  

−(secA−tanA)

Multiplying by secA+tanA in the numerator and denominator of second term,we get

=  

cosA

1

​  

−  

(secA+tanA)

(secA−tanA)(secA+tanA)

​  

 

=  

cosA

1

​  

−  

(secA+tanA)

sec  

2

A−tan  

2

A

​  

 

=  

cosA

1

​  

−  

(secA+tanA)

1

​  

 since sec  

2

A−tan  

2

A=1

=R.H.S

∴  

secA−tanA

1

​  

−  

cosA

1

​  

=  

cosA

1

​  

−  

(secA+tanA)

1

​  

 

Hence proved.