Question

TAN(A+B–C)=1
SEC(B+C–A)=2

· FIND THE VALUES OF A,B AND C
. PLEASE SEND IN WRITTEN​

Answer 1

$\tan(a + b - c) = 1$

tan(a+b-c) =tan 45

a+b-c= 45 ____________(1)

sec(b+c-a) =2

sec(b+c-a) = sec 60

b+c-a= 60 ___________(2)

solve the equation...

a+b-c= 45

-a+b+c= 60

______________

2b= 105

I think now you solve this ques......So,

Please mark as brainliest...