Question

Show that `cot(7 1^circ/2) = sqrt(2) + sqrt(3) + sqrt(4) + sqrt(6)`

Answer 1

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