Question

If A, B, C, D are angles of a cyclic quadrilateral, then prove that i. sin A - sin C = sin D - sin Bii. cos A + cos B + cos C + cos D = 0

Answer 1

If A, B, C, D are angles of a cyclic quadrilateral then, A + C = B + D = 180° ........(1)

i. sinA - sinC = sinD - sinB

LHS = sinA - sinC

= sin(180° - C) - sinC [ from equation (1), ]

= sinC - sinC = 0

RHS = sinD - sinB

= sin(180° - B) - sinB [ from equation (1)]

= sinB - sinB = 0

hence, LHS = RHS


ii. LHS = cosA + cosB + cosC + cosD

= cosA + cos(180° - A) + cosC + cos(180° - C) [ from equation (1), ]

= cosA - cosA + cosC - cosC

= 0 + 0 = 0 = RHS

Answer 2

HELLO DEAR,



GIVEN:-
A, B, C, D are angles of a cyclic quadrilateral then, A + C = B + D = 180° --------- ( 1 )

( i ). sinA - sinC = sinD - sinB

sinA - sinC

=> sin(180° - C) - sinC [ from ------- ( 1 ), ]

=> sinC - sinC = 0

AND
sinD - sinB

=> sin(180° - B) - sinB [ from ---------( 1 )]

=> sinB - sinB = 0

hence, LHS = RHS


( ii ). LHS = cosA + cosB + cosC + cosD

=> cosA + cos(180° - A) + cosC + cos(180° - C) [ from -----------( 1 ) ]

=> cosA - cosA + cosC - cosC

=> 0 + 0 = 0



I HOPE IT'S HELP YOU DEAR,
THANKS