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
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