Question

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

Answer 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.

Answer 2

$\huge\mathcal{Lets \:Assume:-}$

$\implies{ABCD \:is \:a \:cyclic \:quadrilateral}$

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

In quadrilateral $\mathcal\red{ABCD}$

A and C are opposite angles,

$\implies{A + C = 180}$

$\implies{A = 180 - C}$

Similarly, B and D are opposite angles,

$B + D = 180$

$\implies{D = 180 - B}$

$\huge\mathcal\red{L.H.S}$

$cos(180+A)+cos(180-B)+cos(180-C)-sin(90-D)$

$\implies{cos(180+A)+cos(D)+cos(A)-sin(90-D)}$

$\implies{- cos A + cos D + cos A - cos D}$ ( Because, cos (180±x) = - cos x and sin (90 - x) = cos x )

$\implies{0 = R.H.S}$

$\huge\mathcal\red{Hence \:Proved}$