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प्रश्न
Show that `tan(1/2 sin^-1 3/4) = (4 - sqrt(7))/3` and justify why the other value `(4 + sqrt(7))/3` is ignored?
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उत्तर
We have `tan(1/2 sin^-1 3/4)`
Let `1/2 sin^-1 3/4` = θ
⇒ `sin^-1 3/4` = 2θ
⇒ sin 2θ = `3/4`
⇒ `(2 tan theta)/(1 + tan^2theta) = 3/4`
⇒ `3 tan theta^2 - 8` and θ ++ 3 = 0
⇒ tan θ = `(8 +- sqrt(64 - 36))/6`
⇒ tan θ = `(8 +- sqrt(28))/6 = (8 +- sqrt(7))/6 = (4 + sqrt(7))/3`
Now `- pi/2 ≤ sin^-1 3/4 ≤ pi/2`
⇒ `(-pi)/2 ≤ 1/2 sin^-1 3/4 ≤ pi/2`
∴ `tan((-pi)/2) ≤ tan(1/2(sin^-1 3/4)) ≤ tan pi/4`
⇒ `-1 ≤ tan (1/2 sin^-1 3/4) ≤ 1`
⇒ tan θ = `(4 - sqrt(7))/3` ....`(tan theta = (4 + sqrt(7))/3 > 1, "which is not possible")`
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