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SSC CPO 2024 PYQ — Question 139

SSC CPO - 28 June 2024 - Shift 1 • Quantitative Aptitude • Q139

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If (secA+tanA)secA−tanA=25179\frac{(secA + tanA)}{secA - tanA} = 2\frac{51}{79} , then the value of sinA is equal to:

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Correct answer: B

Explanation

The correct answer is option 2 i.e. 65144\frac{65}{144}

sec⁡A+tan⁡Asec⁡A−tan⁡A=25179\frac{\sec A + \tan A}{\sec A - \tan A} = 2\frac{51}{79}

 ⇒ sec⁡A+tan⁡Asec⁡A−tan⁡A=20979\frac{\sec A + \tan A}{\sec A - \tan A} = \frac{209}{79}

⇒ (sec⁡A+tan⁡A)×79=(sec⁡A−tan⁡A)×209(\sec A + \tan A) \times 79 = (\sec A - \tan A) \times 209

⇒ 79sec⁡A+79tan⁡A=209sec⁡A−209tan⁡A79 \sec A + 79 \tan A = 209 \sec A - 209 \tan A

⇒ 79sec⁡A−209sec⁡A=−209tan⁡A−79tan⁡A79 \sec A - 209 \sec A = -209 \tan A - 79 \tan A

⇒ −130sec⁡A=−288tan⁡A-130 \sec A = -288 \tan A

⇒ sec⁡Atan⁡A=288130=14465\frac{\sec A}{\tan A} = \frac{288}{130} = \frac{144}{65}

⇒ sec⁡A=1cos⁡Aandtan⁡A=sin⁡Acos⁡A\sec A = \frac{1}{\cos A} \quad \text{and} \quad \tan A = \frac{\sin A}{\cos A}
⇒ 1cos⁡Asin⁡Acos⁡A=14465\frac{\frac{1}{\cos A}}{\frac{\sin A}{\cos A}} = \frac{144}{65}

⇒  1sin⁡A=14465\frac{1}{\sin A} = \frac{144}{65}

⇒ sin⁡A=65144\sin A = \frac{65}{144}