Question

$\red{EXPLANATION REQUIRED}$

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Answer 1

$\Large{\underline{\underline{\mathfrak{\green{\bf{Solution}}}}}}$

$\Large{\underline{\sf{\:Given}}}$

• Quadratic equation mx² - nx + k = 0
• $\sf{\:\tan 33^{\circ}\:and\:\tan 12^{\circ}\:are\:zeroes}$

$\Large{\underline{\sf{\:Find}}}$

• $\sf{\:Value\:of\:\dfrac{(2m+n+k)}{m}}$

$\Large{\underline{\underline{\mathfrak{\green{\bf{Explanation}}}}}}$

we know,

$\small\boxed{\boxed{\sf{\pink{\:Sum\:of\:zeroes\:=\:\dfrac{-(Coefficient\:of\:x)}{(coefficient\:of\:x^2)}}}}}$

$\implies\sf{\:\tan 33^{\circ}\:+\:\tan 12^{\circ}\:=\:\dfrac{-(-n)}{m}}$

$\implies\sf{\:\tan 33^{\circ}\:+\:\tan 12^{\circ}\:=\:\dfrac{n}{m}.......(1)}$

Again,

$\small\boxed{\boxed{\sf{\pink{\:product\:of\:zeroes\:=\:\dfrac{(Constant\:part)}{(coefficient\:of\:x^2)}}}}}$

$\implies\sf{\:\tan 33^{\circ}\times \tan 12^{\circ}\:=\:\dfrac{k}{m}.......(2)}$

Important formula

★ $\sf{\:\tan(x+y)\:=\:\dfrac{(\tan x + \tan y)}{(1-\tan x.\tan y)}}$

★ $\sf{\:\tan 45^{\circ}\:=\:1}$

So,

$\mapsto\sf{\:\tan (33^{\circ}+12^{\circ})\:=\:\dfrac{(\tan 33^{\circ}+\tan 12^{\circ})}{(1-\tan 33^{\circ} \tan 12^{\circ})}}$

$\mapsto\sf{\:\tan (45^{\circ})\:=\:\dfrac{(\tan 33^{\circ}+\tan 12^{\circ})}{(1-\tan 33^{\circ} .\tan 12^{\circ})}}$

$\mapsto\sf{\:1\:=\:\dfrac{(\tan 33^{\circ}+\tan 12^{\circ})}{(1-\tan 33^{\circ} .\tan 12^{\circ})}}$

$\mapsto\sf{\:(\tan 33^{\circ}\:+\:\tan 12^{\circ})\:=\:(1-\tan 33^{\circ}.\tan 12^{\circ})......(3)}$

Now,

$\implies\sf{\:\dfrac{(2m+n+k)}{m}}$

$\implies\sf{\:\dfrac{2\cancel{m}}{\cancel{m}}\:+\:\dfrac{n}{m}\:+\:\dfrac{k}{m}}$

$\implies\sf{\:\dfrac{n}{m}\:+\:\dfrac{k}{m}+2}$

$\:\:\:\:\small\sf{\:keep\:value\:by\:equ(1)\:and\:(2)}$

$\implies\sf{\:(\tan 33^{\circ}\:+\:\tan 12^{\circ})\:+\:(\tan 33^{\circ}.\tan 12^{\circ})\:+\:2}$

Again,

$\:\:\:\:\small\sf{\:keep\:value\:by\:equ(3)}$

$\implies\sf{(\:1-\cancel{\tan 33^{\circ} \tan 12^{\circ}})+\:\cancel{\tan 33^{\circ}. \tan 12^{\circ}}\:+\:2}$

$\implies\mathfrak{\bf{\:3\:\:\:\:\:\:Ans}}$