Question

If ABCD is a cyclic quadrilateral then the value of cos^2A-cos^2B-cos^2C+cos^2D is 80 minus theta

Answer 1

Given :  ABCD is a cyclic quadrilateral and we know in cyclic quadrilateral opposite angles are supplementary , So

A +  C  =  180°  and B  +  D  =  180°                                  --- ( 1 ) 

And we find value of Cos2 A - Cos2 B - Cos2 C + Cos2 D , As :

We know :  Sin2 θ + Cos2 θ =  1  , So Cos2 θ =  1 - Sin2 θ , Then we use this identity in above equation and get

⇒1 - Sin2 A - ( 1 - Sin2 B  ) - Cos2 C + Cos2 D

⇒1 - Sin2 A - 1 + Sin2 B  - Cos2 C + Cos2 D

⇒- Sin2 A + Sin2 B - Cos2 C + Cos2 D

⇒- ( Sin2 A + Cos2 C  ) + ( Sin2 B  + Cos2 D )


⇒- ( Sin2 ( 180° - C ) + Cos2 C  ) + ( Sin2 ( 180° - D )  + Cos2 D )           ( From equation 1 )

⇒- ( Sin2 C+ Cos2 C  ) + ( Sin2 D  + Cos2 D )                                            ( We know Sin ( 180° - θ ) =  Sin θ  )

⇒- 1 + 1

⇒0