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JEE Main202229 Jun 2022Morning ShiftMathematicsInverse Trigonometric FunctionsActual

50 tan 3 tan - 1 1 2 + 2 cos - 1 1 5 + 4 2 tan 1 2 tan - 1 2 2 is equal to ______.

Correct answer

0

Step-by-step solution

Let S = 50 tan 3 tan - 1 1 2 + 2 cos - 1 1 5 + 4 2 tan 1 2 tan - 1 2 2 = 50 tan tan - 1 1 2 + 2 tan - 1 1 2 + tan - 1 2 + 4 2 tan 1 2 tan - 1 2 2 Let 1 2 tan - 1 2 2 = α ⇒ tan 2 α = 2 2 Now 2 tan α 1 - tan 2 α = 2 2 Solving we get, tan α = 1 2 So S = 50 tan tan - 1 1 2 + 2 · π 2 + 4 2 × 1 2 = 50 tantan - 1 1 2 + 4 = 25 + 4 = 29

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