Question

k=(1-sinA)(1-sin B)(1-sin C)=(1+sin A)(1-sin B)(1-sin C) Find k​

Answer 1

given

k= (1-sinA)(1-sinB)(1-sinC)=(1+sinA)(1+sinB)(1-sinC)

dividing we have

(1+sinA) / (1-sinA)=1

=> 1+ sinA= 1- sin A

=> sinA+ sinA= 0

sinA=0

therefore

k= (1-sinB)(1-sinC)

ANSWER

ANOTHER METHOD

comparison method

we see that

1-sinA= 1+sinA=> 2 sinA=0

and then subtracting the value of

=> sinA= 0

k = ( 1-sinB)(1-sinC)