Question

Prove the following trigonometric identities. $\frac{1+cos\Theta-sin^{2}\Theta }{sin\Theta(1+cos\Theta)} =cot\Theta$

Answer 1

Answer with Step-by-step explanation:

Given :  

(1 + cos θ − sin²θ) / sinθ(1 + cosθ) = cot θ

LHS :   = (1 + cos θ − sin²θ) / sinθ(1 + cosθ)  

= [1 + cosθ − (1+ cos²θ)] / sinθ(1 + cosθ)

[By using  an identity, sin²θ = (1- cos²θ)]

= [1 + cosθ − 1 + cos²θ] / sinθ(1+cosθ)

= [ cosθ + cos²θ)] / sinθ(1+cosθ)

= cosθ(1 + cosθ)] / sinθ(1 + cosθ)

= cosθ/sinθ

= cotθ

(1 + cos θ − sin²θ) / sinθ(1+ cosθ) = cot θ

L.H.S = R.H.S  

Hence Proved..

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

According to the Question :

» (1 + cos θ − sin²θ) / sinθ(1 + cosθ) = cot θ

Where,

LHS : » (1 + cos θ − sin²θ) / sinθ(1 + cosθ)

» [1 + cosθ − (1+ cos²θ)] / sinθ(1 + cosθ)

Note: Using Identity - sin²θ = (1- cos²θ)

» [1 + cosθ − 1 + cos²θ] / sinθ(1+cosθ)

» [ cosθ + cos²θ)] / sinθ(1+cosθ)

» cosθ(1 + cosθ)] / sinθ(1 + cosθ)

» cosθ/sinθ ⇒ cotθ

Also,

» (1 + cos θ − sin²θ) / sinθ(1+ cosθ) = cot θ

∴ LHS = RHS _____[VERIFIED]