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Evaluate the following: tan-13-sec-1(-2) - Mathematics and Statistics

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प्रश्न

Evaluate the following:

`tan^-1 sqrt(3) - sec^-1 (-2)`

योग
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उत्तर

Let `tan^-1(sqrt(3)) = α, "where"  (-pi)/(2) < α < pi/(2)`

∴ tan α = √3 = tan  `pi/(3)`

∴ α = `pi/(3)                    ...[∵ (-pi)/(2) < pi/(3) < pi/(2)]`

∴ `tan^-1(√3) = pi/(3)`        ...(1)

Let sec-1(– 2) = β, where 0 ≤ β ≤ π, β ≠ `pi/(2)`

∴ sec β = – 2 = `- sec  (pi)/(3)`

∴ sec β = `sec(pi - pi/3)`   ...[∵ sec(π – θ) = – secθ]

∴ sec β = `sec  (2pi)/(3)`

∴ β = `(2pi)/(3)                 ...[∵ 0 ≤ (2pi)/(3) ≤ pi]`

∴ sec– 1(– 2) = `(2pi)/(3)`              ...(2)

∴ `tan^-1(√3) - sec^-1(-2)`

= `pi/(3) - (2pi)/(3)`              ...[By (1) and (2)]

= `-(pi)/(3)`.

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अध्याय 3: Trigonometric Functions - Exercise 3.3 [पृष्ठ १०२]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Exercise 3.3 | Q 2.3 | पृष्ठ १०२

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