ii) cos(180° – A) + cos (180° + B) + cos(180° + C)
– sin(90° + D) = 0
friend please answer fast is urgent​

2 thoughts on “ii) cos(180° – A) + cos (180° + B) + cos(180° + C)<br />– sin(90° + D) = 0<br />friend please answer fast is urgent​”

1. Step-by-step explanation:

   A,B,C and D are the angle of a cyclic quadrilateral

   ∴A+C=180

   and B+D=180

   ⇒A=180−C and B=180−D. …(1)

   we have to prove :

   cos(180−A)+cos(180 +B)+cos(180 +C)−sin(90 +D)=0

   LHS=cos(180 −A)+cos(180 +B)+cos(180 +C)−sin(90 +D)

   = −cosA+[−cosB]+[−cosC]−cosD [∵cos(180−θ)=−cosθ,sin(90 +D)=cosD]

   =−cosA−cosB−cosC−cosD

   =−cos(180 −C)−cos(180 −D)−cosC−cosD [from (1)]

   =−(−cosC)−(−cosD)−cosC−cos(D)

   =cosC+cosD−cosD−cosC

   =0

   =RHS

   ∴ cos(180 −A)+cos(180 +B)+cos(180 +C)−sin(90 +D)=0

   Hence proved.

   Reply

2. [tex]\huge\mathcal{Lets \:Assume:-}[/tex]

   [tex]\implies{ABCD \:is \:a \:cyclic \:quadrilateral}[/tex]

   ➡Since, the sum of the opposite angles of a cyclic quadrilateral is supplementary,

   In quadrilateral [tex]\mathcal\red{ABCD}[/tex]

   A and C are opposite angles,

   [tex]\implies{A + C = 180}[/tex]

   [tex]\implies{A = 180 – C}[/tex]

   Similarly, B and D are opposite angles,

   [tex]B + D = 180[/tex]

   [tex]\implies{D = 180 – B}[/tex]

   [tex]\huge\mathcal\red{L.H.S}[/tex]

   [tex]cos(180+A)+cos(180-B)+cos(180-C)-sin(90-D)[/tex]

   [tex]\implies{cos(180+A)+cos(D)+cos(A)-sin(90-D)}[/tex]

   [tex]\implies{- cos A + cos D + cos A – cos D}[/tex] ( Because, cos (180±x) = – cos x and sin (90 – x) = cos x )

   [tex]\implies{0 = R.H.S}[/tex]

   [tex]\huge\mathcal\red{Hence \:Proved}[/tex]

   Reply