If tan x = \(−\frac{12}{5}\), where x lies in the second quadrant, what is the value of sin x − cot x?

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SSC CGL 2023 Tier-I Official Paper (Held On: 24 Jul 2023 Shift 2)
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  1. \(\frac{209}{156}\)
  2. \(\frac{169}{156}\)
  3. \(\frac{156}{209}\)
  4. \(\frac{144}{169}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{209}{156}\)
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PYST 1: SSC CGL - English (Held On : 11 April 2022 Shift 1)
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Detailed Solution

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Given

tan x = -12/5

Concept

tan x = Perpendicular/base

Sin x = Perpendicular/hypotenuse

Calculation

tan x = -12/5 = Perpendicular/base

P = 12 and B = - 5

H2 = P2 + B2

H2 = 144 + 25

H2 = 169

H = 13

⇒ sin x − cot x

⇒ P/H + B/P   ...(since cot is also negative in second quadrant)

⇒ (P2 + BH)PH

⇒ (144 + 65)/156

⇒ 209/156

The value is 209/156.

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