If \(\rm \frac{(\sec A+\tan A)}{\sec A-\tan A}=2\frac{51}{79}\) then the value of sin A is equal to:

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This question was previously asked in

SSC CPO 2024 Official Paper-I (Held On: 28 Jun, 2024 Shift 1)

Answer 1

Given :-

\(\rm \frac{(\sec A+\tan A)}{\sec A-\tan A}=2\frac{51}{79}\)

Formula Used :-

secθ = \(\frac{1}{cosθ}\)

tanθ = \(\frac{sin\theta}{cos\theta}\)

Calculation :-

\(\rm \frac{(\sec A+\tan A)}{\sec A-\tan A}=2\frac{51}{79}\)

⇒ \(\frac{\frac{1}{cosA} + \frac{sinA}{cosA}}{\frac{1}{cosA} - \frac{sinA}{cosA}} = \frac{209}{79}\)

⇒ \(\frac{1 + sinA}{1 - sinA} = \frac{209}{79}\)

⇒ 209 - 209sinA = 79 + 79sinA

⇒ 209 - 79 = 209sinA + 79sinA

⇒ 130 = 288sinA

⇒ sinA = \(\frac{130}{288}\)

⇒ sinA = \(\frac{65}{144}\)