Question

pls answer this urgently​

Answer 1

Hope it helps :)

Ans. is in the attachment

Answer 2

GIVEN:

sinθ + cosθ // sinθ - cosθ  +  sinθ - cosθ // sinθ + cosθ

TO PROVE:

2 //  2sin^2 θ - 1

PROOF:

sinθ + cosθ // sinθ - cosθ  + sinθ - cosθ // sinθ + cosθ

{// ⇒ whole divide }

(sinθ + cosθ)^2  +  (sinθ - cosθ)^2    //    sin^2 θ  -  cos^2 θ                        {^2 ⇒ square}

2 (sinθ + cosθ)^2    //     sin^2 θ - cos^2 θ

WE KNOW THAT,

(sin^2 θ + cos^2 θ  =  1      &   cos^2 θ = 1 - sin^2 θ

SUBSTITUTING THESE WE GET,

2 //  sin^2 θ - (1 - sin^2 θ)

2 //  2sin^2 θ - 1

HENCE PROVED ^_^