Q53 of 268 Page 4

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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