Question

If A+B+C=90° , prove that sin²A+sin²B+sin²C=1-2sinA sinB sinC

Answer 1

ans: ELABORATE METHOD:-

C = 90 - A - B

sin2A+sin2B+cos2(A+B)+2sinAsinBcos(A+B)

= sin2A+sin2B+(cosAcosB−sinAsinB)2+2sinAsinB(cosAcosB−sinAsinB)

=sin2A+sin2B+(cos2Acos2B+sin2Asin2B−2sinAsinBcosAcosB)+(2sinAsinBcosAcosB−2sin2Asin2B)

=sin2A+sin2B+cos2Acos2B−sin2Asin2B

=sin2A(1−sin2B)+sin2B+cos2Acos2B

=sin2Acos2B+sin2B+cos2Acos2B

=cos2B(sin2A+cos2A)+sin2B

=cos2B+sin2B

= 1

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