# If tan A + sec A = 3, find the value of sin A.

tan A + sec A = 3

tanA = 3 secA

Squaring both the sides, we get

tan2 A =(3 – secA)2

tan2 A = 9 + sec2A – 6sec A

sec2 A – 1 = 9 + sec2A – 6sec A [ 1+ tan2 A = sec2 A]

–1 – 9 = –6secA

– 10 = –6sec A

Now, tan A + sec A = 3

sin A = 3cosA 1

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