Cos^2a-cos^2b-cos^2c+cos^2d. When abcd cyclic quadrilateral
Answers
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
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