Question

prove that tan theta tan(60 + theta)tan(60 - theta) = tan 3 theta and hence deduce that tan 20° tan 40° tan 60° tan 80° = 3. I proved the first part but I need help with deducing the values

Answer 1

HELLO DEAR,

we know:-

tan(A - B) = {tanA - tanB} / (1 + tanAtanB)-----( 1 )

tan(A + B) = {tanA + tanB} / (1 - tanAtanB)-----( 2 )

multiply--------( 1 ) & ------( 2 )

tan(A - B)*tan(A + B) = {tanA - tanB} / (1 + tanAtanB) * {tanA + tanB} / (1 - tanAtanB)

tan(A - B)*tan(A + B) = (tan²A - tan²B) / (1 - tan²A tan²B)

put A = 60° and B = ${\Theta}$ we get,

tan(60 - ${\Theta}$)tan(60 + ${\Theta}$) = $\bold{\frac{(\sqrt{3})^2 - tan^2{\Theta}}{1 - (\sqrt{3})^2tan^2{\Theta}}}$

tan(60 - ${\Theta}$)tan(60 + ${\Theta}$) = $\bold{\frac{3 - tan^2{\Theta}}{1 - 3tan^2{\Theta}}}$

tan(60 - ${\Theta}$)tan(60 + ${\Theta}$) = $\frac{tan\Theta}{tan\Theta}$ * $\bold{\frac{3 - tan^2{\Theta}}{1 - 3tan^2{\Theta}}}$

tan(60 - $\Theta$)tan(60 + $\Theta$) = $\bold{\frac{1}{tan\Theta}}$ * $\bold{\frac{3tan\Theta - tan^3\Theta}{1 - 3tan^2\Theta}}$

we know :- tan3A = $\bold{\frac{3tan\Theta - tan^3\Theta}{1 - 3tan^2\Theta}}$

tan$\Theta$tan(60 - $\Theta$)tan(60 + $\Theta$) = tan3$\Theta$----------( 3 )

put $\Theta$ = 20, in ---------( 3 )

tan(20) tan(60 - 20) tan(60 + 20) = tan(3*20)

multiply both side by "tan60"

tan(20) tan(40) tan(60) tan(80) = tan(60) * tan(60)

tan(20) tan(40) tan(60) tan(80) = (√3) * (√3)

tan(20) tan(40) tan(60) tan(80) = 3

$\bold{\Large{HENCE, \,\,PROVED}}$

I HOPE ITS HELP YOU DEAR,T

THANKS

Answer 2

Answer:

Please mark me brainliest