Question

A traiangle ABC is right angle triangle at b find the value of sec A, sin B-10A 10 C by sinB​

Answer 1

Heyy,

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

❤️

Answer 2

Answer:

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