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प्रश्न
Evaluate the following:
`tan^-1(1) + cos^-1(1/2) + sin^-1(1/2)`
उत्तर
Let tan- 1(1) = α, where `(-pi)/(2) < α < pi/(2)`
∴ tan α = 1 = `tan pi/(4)`
∴ α = `pi/(4) ...[∵ (-pi)/(2) < pi/(4) < pi/(2)]`
∴ tan– 1(1) = `pi/(4)` ...(1)
Let `cos^-1(1/2)` = β, where 0 ≤ β ≤ π
∴ cos β = `1/2 = cos (pi)/(3)`
∴ β = `pi/(3) ...[∵ 0 < pi/(3) < pi]`
∴ `cos^-1(1/2) = pi/(3)` ...(2)
Let `sin^-1(1/2) = γ, "where" (-pi)/(2) ≤ γ ≤ pi/(2)`
∴ sin γ = `(1)/(2) = sin (pi)/(6)`
∴ γ = `pi/(6) ...[∵ (-pi)/(2) ≤ pi/(6) ≤ pi/(2)]`
∴ `sin^-1(1/2) = pi/(6)` ...(3)
∴ `tan^-1(1) + cos^-1(1/2) + sin^-1(1/2)`
= `pi/(4) + pi/(3) + pi/(6)` ...[By (1), (2) and (3)]
= `(3pi + 4pi + 2pi)/(12)`
= `(9pi)/(12)`
= `(3pi)/(4)`.
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