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

Asked by Omsingh5701 | 10th Dec, 2019, 09:59: PM

Expert Answer:

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

Answered by Sneha shidid | 11th Dec, 2019, 09:45: AM