Question

sinø/1+cosø + 1+cosø/sinø = 2 cosecø​

Answer 1

Step-by-step explanation:

let's assume theta as A,

by LHS,

sinA/(1+cosA) + (1+cosA)/sinA

=sin^2 A + (1+cosA)^2/ sinA(1+cosA)

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

=1 + 1 + 2cosA / sinA(1+cosA)

=2( 1 + cosA)/sinA (1+cosA)

=2/sinA

=2cosecA

Answer 2

It is true that sinθ/(1+cosθ) + (1+cosθ)/sinθ = 2 cosecθ

Given:

sinθ/(1+cosθ) + (1+cosθ)/sinθ = 2 cosecθ

To find:

Prove that sinθ/(1+cosθ) + (1+cosθ)/sinθ = 2 cosecθ

Solution:

Given that sinθ/(1+cosθ) + (1+cosθ)/sinθ = 2 cosecθ  

⇒ Take LHS and simplify

= sinθ/(1+cosθ) + (1+cosθ)/sinθ

= [sin²θ + (1+cosθ)²] /sinθ(1+cosθ)      [ take sinθ(1+cosθ) as LCM ]

as we know   (a+b)²= a²+b²=2ab  

⇒  (1+cosθ)² = 1 + cos²θ +2cosθ  

= [sin²θ +1 + cos²θ +2cosθ] /sinθ(1+cosθ)  

= [ sin²θ + cos²θ + 1 +2cosθ]/ sinθ(1+cosθ)    

= [1 + 1 + 2cosθ] / sinθ(1+cosθ)  

= [2+2cosθ] / sinθ(1+cosθ)    

= [2(1+cosθ)] / sinθ(1+cosθ)    

= 2/sinθ

= 2 (1/sinθ)  

As we know 1/sinθ = cosecθ

= 2cosecθ  = RHS

Therefore, LHS = RHS

Hence,

It is proven that sinθ/(1+cosθ) + (1+cosθ)/sinθ = 2 cosecθ

#SPJ2