ong>ANSWER:

We have to prove that

cot+cosecθ

sinθ

​  

=2+  

cotθ−cosecθ

sinθ

​  

or,  

cotθ+cosecθ

sinθ

​  

−  

cotθ−cosecθ

sinθ

​  

=2

Now,

⇒ LHS =  

cotθ+cosecθ

sinθ

​  

−  

cotθ−cosecθ

sinθ

​  

⇒ LHS =  

cosecθ+cotθ

sinθ

​  

+  

cosecθ−cotθ

sinθ

​  

⇒ LHS = sinθ{  

cosecθ+cotθ

1

​  

+  

cosecθ−cotθ

1

​  

}

⇒ LHS = $$sin \, \theta\left\{\dfrac{cosec \, \theta \, - \, cot \, \theta \, + \, cosec \, \theta \, + \, cot \, \theta}{cosec^2 \, \theta \, - \, cot^ \, \theta}\right\} \, = \, sin \, \theta \, \left(\dfrac{2 \, cosec \, \theta}{1} \right)$$

⇒ LHS = sinθ(2cosecθ)=2sinθ×  

sinθ

1

​  

=2=RHS

⇒ LHS = 2 = RHS

ALTERNANATIVELY,

LHS =  

cotθ+cosecθ

sinθ

​  

⇒LHS=sinθ(cosecθ−cotθ)      [∵  

cosecθ+cotθ

1

​  

=cosecθ−cotθ]

⇒ LHS = sinθ(  

sinθ

1

​  

−  

sinθ

cosθ

​  

)=sinθ(  

sinθ

1−cosθ

​  

)

⇒ LHS = 1 - COS θ

⇒ = 2 - (1 + cos θ)

⇒ LHS = 2 -  

1−cosθ

(1+cosθ)(1−cosθ)

​  

⇒ LHS = 2 -  

1−cosθ

(1−cos  

2

θ)

​  

⇒ LHS = 2 -  

1−cosθ

sin  

2

θ

​  

=2−  

sinθ

1−cosθ

​  

sinθ

​  

=2−  

sinθ

1

​  

−  

sinθ

cosθ

​  

sinθ

​  

⇒ LHS = 2 -  

cosecθ−cotθ

sinθ

​  

 = RHS

Hope it help's you

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