Show that # tan(52.5°) = sqrt6 − sqrt3 − sqrt2 + 2# ?

c3 maths question

3 Answers
Abhishek K.
Mar 1, 2018

#rarrtan75°=tan(45+30)#

#=(tan45+tan30)/(1-tan45*tan30)#

#=(1+(1/sqrt(3)))/(1-(1/sqrt(3))#

#=(sqrt(3)+1)/(sqrt(3)-1)=2+sqrt(3)#

#rarrtan52.5=cot(90-37.5)=cot37.5#

#rarrcot37.5=1/(tan(75/2))#

#rarrtanx=(2tan(x/2))/(1-tan^2(x/2))#

#rarrtanx-tanx*tan^2(x/2)=2tan(x/2)#

#rarrtanx*tan^2(x/2)+2tan(x/2)-tanx=0#

It is quadratic in #tan(x/2)# So,

#rarrtan(x/2)=(-2+sqrt(2^2-4*tanx*(-tanx)))/(2*tanx)#

#rarrtan(x/2)=(-2+sqrt(4(1+tan^2x)))/(2*tanx)#

#rarrtan(x/2)=(-1+sqrt(1+tan^2x))/tanx#

Putting #x=75# we get

#rarrtan(75/2)=(-1+sqrt(1+tan^2(75)))/(tan75)#

#rarrtan(75/2)=(-1+sqrt(1+(2+sqrt(3))^2))/(2+sqrt(3))#

#rarrtan(75/2)=(-1+sqrt(1+4+4sqrt(3)+3))/(2+sqrt(3))#

#rarrtan(75/2)=(-1+sqrt(8+4sqrt(3)))/(2+sqrt(3))#

#rarr1/tan(75/2)=(2+sqrt(3))/(2*sqrt(2+sqrt(3))-1)*(2*sqrt(2+sqrt(3))+1)/(2*sqrt(2+sqrt(3)+1)#

#rarrcot37.5=(2*(2*sqrt(2+sqrt(3))+1)+sqrt(3)*(2*sqrt(2+sqrt(3))+1))/((2*sqrt(2+sqrt(3)))^2-1^2)#

Let #sqrt(2+sqrt(3))=a#

#rarrcot37.5=(2*(2*a+1)+sqrt(3)*(2*a+1))/((4*(2+sqrt(3)))^2-1)#

#rarrcot37.5=(4a+2+2sqrt(3)a+sqrt(3))/((4*(2+sqrt(3))-1)#

#rarrcot37.5=(4a+2sqrt(3)a+a^2)/(7+4sqrt(3))*(7-4sqrt(3))/(7-4sqrt(3))#

#rarrcot37.5=7*(4a+2sqrt(3)a+a^2)-4sqrt(3)*(4a+2sqrt(3)a+a^2)#

#rarrcot37.5=28a+14sqrt(3)a+7a^2-16sqrt(3)a-24a-4sqrt(3)a^2#

#rarrcot37.5=7a^2-4sqrt(3)a^2+4a-2sqrt(3)a#

#rarrcot37.5=a^2(7-4sqrt(3))+2*a(2-sqrt(3))#

#rarrcot37.5=(2+sqrt(3))(7-4sqrt(3))+2*sqrt(2+sqrt(3))*sqrt((2-sqrt(3)))*sqrt((2-sqrt(3)))#

#rarrcot37.5=2*(7-4sqrt(3))+sqrt(3)(7-4sqrt(3))+sqrt(2^2*(2-sqrt(3)))#

#rarrcot37.5=14-8sqrt(3)+7sqrt(3)-12+sqrt((sqrt(6)-sqrt(2))^2)#

#rarrtan52.5=2-sqrt(3)+sqrt(6)-sqrt(2)#

Proved.

Sunayan S
Mar 1, 2018

Smaller approach...

Rules Used:-

#color(red)(ul(bar(|color(green)(sin2theta=2 cdot sintheta cdot costheta))|#

#cos2theta=2cos^2theta-1#

#=>color(red)(ul(bar(|color(blue)(2cos^2theta=1+cos2theta))|#

Explanation:

#tan(52.5^@)#

#=sin(52.5^@)/cos(52.5^@)#

#=sin(105/2)^@/cos(105/2)^@#

#=(2 cdot sin(105/2)^@ cdot cos(105/2)^@) /(2 cdot cos(105/2)^@ cdot cos(105/2)^@#

#=sin(105/2 xx2)^@/(2 cdot cos^2(105/2)^@#

#=sin(105)^@/(cos(105)^@+1)#

#=sin(60^@+45^@)/(cos(60^@+45^@)+1)#

#=(sin60^@ cdot cos45^@+cos60^@ cdot sin45^@)/(cos60^@ cdot cos45^@ -sin60^@ cdot sin45^@+1 #

#=(sqrt3/2 cdot 1/sqrt2+1/2 cdot 1/sqrt2)/(1/2 cdot 1/sqrt2-sqrt3/2 cdot 1/sqrt2+1#

#=((sqrt3+1)/(2sqrt2))/((1-sqrt3+2sqrt2)/(2sqrt2)#

#=(sqrt3+1)/(1-sqrt3+2sqrt2#

#=((sqrt3+1) cdot (1+2sqrt2+sqrt3))/((1+2sqrt2)^2-(sqrt3)^2)#

#=(sqrt3+2sqrt6+3+1+2sqrt2+sqrt3)/(1+4sqrt2+8-3)#

#=(2(sqrt6+sqrt3+sqrt2+2))/(6+4sqrt2)#

#=((sqrt6+sqrt3+sqrt2+2))/(3+2sqrt2)#

#=((3-2sqrt2)(sqrt6+sqrt3+sqrt2+2))/((3+2sqrt2)(3-2sqrt2)#

#=(sqrt6-sqrt3-sqrt2+2)/1#

#=sqrt6-sqrt3-sqrt2+2#

Hope it helps...
Thank you...

P dilip_k
Mar 1, 2018

#tan105^@=tan(60^@+45^@)#
#=>tan105^@=(tan60^@+tan45^@)/(1-tan60^@tan45^@)#

#=>tan105^@=(sqrt3+1)/(1-sqrt3*1)#

#=>tan105^@=(1+sqrt3)/(1-sqrt3)#

#=>tan105^@=-((sqrt3+1)(sqrt3-1))/(sqrt3-1)^2#

#=>tan105^@=-(3-1)/(4-2sqrt3)#

#=>(2tan52.5^@)/(1-tan^2 52.5^@)=-1/(2-sqrt3)#

Let #tan52.5^@=x#

Now

#(2x)/(1-x^2)=-1/(2-sqrt3)#

#=>x^2-2(2-sqrt3)x-1=0#

#=>x=(2(2-sqrt3)+sqrt(4(2-sqrt3)^2+4))/2#

As #52.5^@in"First Quadrant -ve root neglected"#

#=>x=(2(2-sqrt3)+2sqrt((2-sqrt3)^2+1))/2#

#=>x=(2-sqrt3)+sqrt((2-sqrt3)^2+1)#

#=>x=(2-sqrt3)+sqrt(8-4sqrt3)#

#=>x=(2-sqrt3)+sqrt(2(4-2sqrt3)#

#=>x=(2-sqrt3)+sqrt(2(sqrt3-1)^2)#

#=>x=2-sqrt3+sqrt2(sqrt3-1)#

#=>x=2-sqrt3+sqrt6-sqrt2#

#=>x=sqrt6-sqrt3-sqrt2+2#

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