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प्रश्न
Show that `2 cot^(-1) 3/2 + sec^(-1) 13/12 = π/2`
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उत्तर
`2 cot^(-1) 3/2 = 2 tan^(-1) 2/3 ...[because cot^(-1) x = tan^(-1) (1/x)]`
`= tan^(-1) [(2 × 2/3)/(1 - (2/3)^2)] ...[because 2 tan^(-1) x = tan^(-1) ((2x)/(1 - x^2))]`
`= tan^(-1) [(4/3)/(1 - 4/9)]`
`= tan^(-1) (4/3 × 9/5)`
`= tan^(-1) 12/5` ....(1)
Let `sec^(-1) 13/12 = α`
Then, `sec α = 13/12, "where" 0 < α < pi/2`
∴ `tan α > 0`
Now, `tan α = sqrt(sec^2 α - 1)`
`tan α = sqrt(169/144 - 1)`
`tan α = sqrt(25/144)`
`tan α = 5/12`
∴ `α = tan^(-1) 5/12 = cot^(-1) 12/5 .....[because tan^(-1) x = cot^(-1) (1/x)]`
∴ `sec^(-1) 13/12 = cot^(-1) 12/5` .....(2)
Now,
LHS = `2 cot^(-1) 3/2 + sec^(-1) 13/12`
LHS = `tan^(-1) 12/5 + cot^(-1) 12/5` ...[By (1) and (2)]
LHS = `π/2 .....[because tan^(-1) x + cot^(-1) x - π/2]`
RHS = `π/2`
LHS = RHS
`2 cot^(-1) 3/2 + sec^(-1) 13/12 = π/2`
Hence proved.
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