Question

Prove That


$\sf\longrightarrow \sf \ \dfrac{2 - cosec^2A}{cosec^2A + 2cotA} = \dfrac{sinA - cosA}{sinA + cosA}$

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Answer 1

Step-by-step explanation:

Your Answer with explanation

Answer 2

Let, i consider A be θ

(2 - cosec²θ)/(cosec²θ + 2cotθ) = (sinθ - cosθ)/(sinθ + cosθ)

from R.H.S ,

[ 2 - (1/sin²θ) ]/[ (1/sin²θ) + (2cosθ/sinθ) ]

(2sin²θ - 1/sin²θ)/(1/sinθ) + (2cosθ/sinθ)

(2sin²θ - 1)/sin²θ(1/sin²θ + 2cosθ/sinθ)

(2sin²θ - 1)/(1 + 2cosθ•sinθ)

(sin²θ + sin²θ - 1)/( sin²θ + cos²θ + 2cosθ•sinθ)

(sin²θ + cos²θ)/(sinθ + cosθ)²

[ (sinθ + cosθ) • (sinθ - cosθ) ]/(sinθ + cosθ)²

(sinθ - cos)/(sinθ + cosθ)

HENCE PROVED