Question

Prove that tan 11pi/24 =√2+√3+√4+√6

Answer 1

$\large\underline{\sf{Solution-}}$

Given Trigonometric function is

$\rm :\longmapsto\:tan\bigg[\dfrac{11\pi}{24} \bigg]$

$\rm \:  =  \: \:tan\bigg[\dfrac{11 \times 180 \degree \: }{24} \bigg]$

$\rm \:  =  \: \:tan\bigg[\dfrac{11 \times 15 \degree \: }{2} \bigg]$

$\rm \:  =  \: \:tan\bigg[\dfrac{165 \degree \: }{2} \bigg]$

$\rm \:  =  \: \:tan\bigg[82 \: \dfrac{ \: 1 \degree}{ \: 2 \: } \bigg]$

$\rm \:  =  \: \:tan\bigg[90\degree - 7 \: \dfrac{\: 1 \degree}{ \: 2 \: } \bigg]$

$\rm \:  =  \: \:cot\bigg[7 \: \dfrac{\: 1 \degree}{ \: 2 \: } \bigg]$

$\rm \:  =  \: \:cot\bigg[ \: \dfrac{\: 15 \degree}{ \: 2 \: } \bigg]$

Now, we know that

$\red{ \boxed{ \sf{ \:\:cotA = \dfrac{cosA}{sinA} = \dfrac{ {2cos}^{2} A}{2sinAcosA} = \dfrac{1 + cos2A}{sin2A} }}}$

$\rm \:  =  \: \dfrac{1 + cos15\degree }{sin15\degree }$

$\rm \:  =  \: \dfrac{1 + cos(45\degree - 30\degree ) }{sin(45\degree - 30\degree)}$

We know,

$\red{ \boxed{ \sf{ \:cos(x - y) = cosxcosy + sinxsiny}}}$

and

$\red{ \boxed{ \sf{ \:sin(x - y) = sinxcosy - sinycosx}}}$

So, using these Identities, we get

$\rm \:  =  \: \dfrac{1 + cos45\degree cos30\degree + sin45\degree sin30\degree }{sin45\degree cos30\degree - sin30\degree cos45\degree }$

$\rm \:  =  \: \dfrac{1 + \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} + \dfrac{1}{ \sqrt{2} } \times \dfrac{1}{2} }{\dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2} \times \dfrac{1}{ \sqrt{2} } }$

$\rm \:  =  \: \dfrac{1 + \dfrac{ \sqrt{3} }{2 \sqrt{2} } + \dfrac{1}{2 \sqrt{2} } }{\dfrac{ \sqrt{3} }{2 \sqrt{2} } - \dfrac{1}{2 \sqrt{2} } }$

$\rm \:  =  \: \dfrac{2 \sqrt{2} + \sqrt{3} + 1}{ \sqrt{3} - 1 }$

On rationalizing the denominator, we get

$\rm \:  =  \: \dfrac{2 \sqrt{2} + \sqrt{3} + 1}{ \sqrt{3} - 1 } \times \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} + 1 }$

$\rm \:  =  \: \dfrac{2 \sqrt{6} + 3 + \sqrt{3} + 2 \sqrt{2} + \sqrt{3} + 1}{ { (\sqrt{3} )}^{2} - {1}^{2} }$

$\rm \:  =  \: \dfrac{2 \sqrt{6} + 4 +2 \sqrt{3} + 2 \sqrt{2}}{ 3 - 1 }$

$\rm \:  =  \: \dfrac{2 \sqrt{6} + 4 +2 \sqrt{3} + 2 \sqrt{2}}{2}$

$\rm \:  =  \: \sqrt{6} + 2 + \sqrt{3} + \sqrt{2}$

$\rm \:  =  \: \sqrt{6} + \sqrt{4} + \sqrt{3} + \sqrt{2}$

Hence,

$\bf :\longmapsto\:tan\bigg[\dfrac{11\pi}{24} \bigg] = \sqrt{6} + \sqrt{4} + \sqrt{3} + \sqrt{2}$