A triangle ABC is right angled at B;find the value of secA. SinC- tanA. TanC /sinB

In triangle ABC,

A + B + C = 180°

A + C = 90°               B = 90°

C = 90° - A

 

(sec A sin C - tan A tan C) /sin B

= (sec A sin (90° - A) - tan A tan (90° - A)) /sin 90°

= (sec A cos A - tan A cot A)/1

= 1 - 1

= 0