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Question
Show that `cot(7 1^circ/2) = sqrt(2) + sqrt(3) + sqrt(4) + sqrt(6)`
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Solution
We have to prove that `cot(7 1^circ/2) = sqrt(2) + sqrt(3) + sqrt(4) + sqrt(6)`
L.H.S = `cot(7 1^circ/2)`
= `(cos(7 1^circ/2))/(sin(7 1^circ/2))`
To find `costheta/sintheta`, multiply numerator and denominator by 2 cos θ
Let θ = `71/2^circ`
2θ = 15°
`(2cos^2theta)/(2sin theta cos theta) = (1 + cos 2theta)/(sin 2theta)`
= `(1 + cos 15^circ)/(sin 15^circ)`
= `((1 + sqrt(3) + 1)/(2sqrt(2)))/((sqrt(3) - 1)/(2sqrt(2))`
= `(2sqrt(2) + sqrt(3) + 1)/(sqrt(3) - 1)`
Multiply numerator and denominator by `sqrt(3) + 1`
= `((2sqrt(2) + sqrt(3) + 1)(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1))`
= `(2sqrt(2) + 3 + sqrt(3) + 1)/(3 - 1)`
= `(2sqrt(3) + 2sqrt(2) + 4)/2`
= `(2(sqrt(2) + sqrt(3) + sqrt(6) + 2))/2`
= `sqrt(2) + sqrt(3) + sqrt(4) + sqrt(6)`
= R.H.S
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