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प्रश्न
Find the value of tan22°30′. `["Hint:" "Let" θ = 45°, "use" tan theta/2 = (sin theta/2)/(cos theta/2) = (2sin theta/2 cos theta/2)/(2cos^2 theta/2) = sintheta/(1 + costheta)]`
उत्तर
Let 22°30′ = `theta/2`
∴ θ = 45°
tan22°30′ = `tan theta/2`
= `(sin theta/2)/(cos theta/2)`
= `(2sin theta/2 cos theta/2)/(2cos^2 theta/2)`
= `sintheta/(1 + costheta)`
Put θ = 45°
∴ `sintheta/(1 + costheta) = sin 45^circ/(1 + cos 45^circ)`
= `(1/sqrt(2))/(1 + 1/sqrt(2))`
= `1/(sqrt(2) + 1)`
= `(1 xx (sqrt(2) - 1))/((sqrt(2) + 1)(sqrt(2) - 1))`
= `sqrt(2) - 1`
Hence, tan22°30' = `sqrt(2) - 1`.
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