Question

if x cos aa+y sin a=x cos b+y sin b =k show that x/cos 1/2(a+b)=y/sin 1/2(a-b)​

Answer 1

Answer:

☆☺️✨✨♡ANSWER ☺️✨✨✨☆

Step-by-step explanation:

Correct option is

Correct option isA

Correct option isAcosα+cosβ=x2+y24ax

Correct option isAcosα+cosβ=x2+y24axB

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2C

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ay

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes 

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.Or 

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.Or (ysinθ−2a)2=(−xcosθ)2

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.Or (ysinθ−2a)2=(−xcosθ)2Or 

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.Or (ysinθ−2a)2=(−xcosθ)2Or y2sin2θ−4aysinθ+4a2=x2cos2θ

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.Or (ysinθ−2a)2=(−xcosθ)2Or y2sin2θ−4aysinθ+4a2=x2cos2θOr 

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.Or (ysinθ−2a)2=(−xcosθ)2Or y2sin2θ−4aysinθ+4a2=x2cos2θOr y2sin2θ−4aysinθ+4a2=x2(1−sin2θ)

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.Or (ysinθ−2a)2=(−xcosθ)2Or y2sin2θ−4aysinθ+4a2=x2cos2θOr y2sin2θ−4aysinθ+4a2=x2(1−sin2θ)Or 

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.Or (ysinθ−2a)2=(−xcosθ)2Or y2sin2θ−4aysinθ+4a2=x2cos2θOr y2sin2θ−4aysinθ+4a2=x2(1−sin2θ)Or (x2+y2)sin2θ−4aysinθ+4a2+x2=0.

Correct option isAcosα+cosβ=x2+y24axBcosαcosβ=x2+y24a2−y2Csinα+sinβ=x2+y24ayThe general equation becomes xcosθ+ysinθ=2a.Or (ysinθ−2a)2=(−xcosθ)2Or y2sin2θ−4aysinθ+4a2=x2cos2θOr y2sin2θ−4aysinθ+4a2=x2(1−sin2θ)Or (x2+y2)sin2θ−4aysinθ+4a2+x2=0.Let the roots be