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ExamsJEE MainMaths

Evaluate: (i) sec( tan⁻¹( tan(-pi/3))) (ii) cot( cot⁻¹(1/2)) (iii) tan( cos⁻¹(3/4) + sin⁻¹(3/4) - sec⁻¹(3))

  1. (i) 2; (ii) 1/2; (iii) cot(sec⁻¹ 3) = 1/sqrt(8)
  2. (i) 1/2; (ii) 2; (iii) sqrt(8)
  3. (i) -2; (ii) 1/2; (iii) 0
  4. (i) 2; (ii) 1/2; (iii) tan(sec⁻¹ 3) = sqrt(8)

Correct answer: (i) 2; (ii) 1/2; (iii) cot(sec⁻¹ 3) = 1/sqrt(8)

Solution

Use inverse-function identities. In (i), tan(-pi/3) = -sqrt3 lies in the principal range, so tan⁻¹ returns -pi/3, and sec(-pi/3) = 2. In (ii), cot of cot inverse returns the argument 1/2. In (iii), cos⁻¹(3/4) + sin⁻¹(3/4) = pi/2, so the expression is tan(pi/2 - sec⁻¹ 3) = cot(sec⁻¹ 3) = 1/sqrt(8).

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