Question 35595
If anglesA,B, and C are the sides of a triangle such that sin(A+B)=1/cos(C)
IN A TRIANGLE A+B+C=180.HENCE 
SIN(A+B)=SIN(180-C)=SIN(C)=1/COS(C)
SIN(C)COS(C)=1
SIN(2C)/2=1...
SIN?(2C)=2...IMPOSSIBLE.
IT IS WRONG .SIN(2C) CAN NOT EQUAL 2.IT COULD BE SIN(A+B)=1+COS(C)
THEN AS PROVED ABOVE 
SIN(C)=1+COS(C) WHICH CAN BE SOLVED TO GET C=90.PLEASE CHECK UP AND INFORM

 and cos(A+B)=cos(C),
COS(180-C)= - COS(C)=COS(C)
2COS(C)=0
COS(C)=0
C=90 DEG. HENCE ABC IS RIGHT ANGLED AT C.
 then show that the triangle is a right triangle