in triangle ABC ,right-angled at C, find cos A,tan A and cosec B if sinA =24/25

ABC is a right angled triangle right angled at C = C = 90

Also for a triangle we know that

A + B + C = 180

= A + B + 90 = 180 = A + B = 180 - 90

= A + B = 90

= A = (90 - B) ..........(1)

Given sin A = 24/25

Now sin² A + cos² A = 1

= cos² A = 1 - sin² A = cos² A = 1 - (24/25)² (Putting value of sin A)

= cos² A = 1 - 576/625 = cos² A = (625 - 576)/625

= cos² A = 49/625 = cos A = √(49/625)

= cos A = ±7/25

tan A = sin A / cos A = (24/25) / (±7/25)

= tan A = (24/25)(± 25/7)

= tan A = ± 24/7

cosec B = cosec(90 - A) (Putting value of B from equation (1) )

= sec A = 1/cos A

= 1/(±7/25)

= ±25/7

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