Ex 2.1, 14 - Chapter 2 Class 12 Inverse Trigonometric Functions

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Ex 2.1, 14 Find the value of tan−1 √3 – sec−1(–2) is equal to π (B) – π/3 (C) π/3 (D) 2π/3 Solving tan−1 √𝟑 Let y = tan−1 √3 tan y = √3 tan y = tan (π/3) Range of principal value of tan−1 is ((−π)/2.π/2) Rough We know that tan 60° = √3 θ = 60° = 60 × 𝜋/180 = 𝜋/3 Since √3 is positive Principal value is θ i.e. 𝜋/3 Hence, principal value of tan−1 √3 is π/3 Solving sec−1 (–2) Let y = sec−1 (–2) sec y = −2 sec y = sec (2π/3) We know that principal value range of sec−1 is [0,π] – {π/2} Rough We know that cos 60° = 1/2 sec 60° = 2 θ = 60° = 60 × 𝜋/180 = 𝜋/3 Since −2 is negative Principal value is π – θ i.e. π – 𝜋/3 = 2𝜋/3 Hence, principal value of sec−1 (–2) = 2π/3 Now we have tan−1 (√3) = π/3 & sec−1 (–2) = 2π/3 Solving tan−1 √𝟑 – sec−1 (−2) = π/3 − (2π/3) = π/3 − 2π/3 = (π − 2π)/3 = (−π)/3 Hence , correct answer is (B)