#(sec a+1)=tana(2+√3) # 0<=a<=2π??

2 Answers
P dilip_k
Feb 2, 2018

#(sec a+1)=tana(2+√3) #

#=>(seca+1)xxcosa=cosaxxtana(2+√3) #

#=>(cosa+1)=sina(2+√3) #

#=>(2+√3)2sin(a/2)cos(a/2)=2cos^2(a/2) #

#=>[(2+√3)sin(a/2)-cos(a/2)]2cos(a/2)=0#

So #cos (a/2)=0=cos(pi/2)#

#=>a/2=pi/2#

#=>a=pi#

Again
#[(2+√3)sin(a/2)-cos(a/2)]=0#

#=>(2+√3)sin(a/2)=cos(a/2)]#

#=>sin(a/2)/cos(a/2)=1/(2+√3)#

#=>tan(a/2)=2-sqrt3=tan(pi/12)# #color(red)"see note below"#

#=>a/2=pi/12#

#=>a=pi/6#

Alternative method

#(sec a+1)=tana(2+√3) #

#=>(sec a+1)-tana(2+√3)=0 #

#=>(sqrt(sec a+1))^2-(sqrt(sec^2a-1))(2+√3)=0 #

#=>(sqrt(sec a+1))[sqrt(seca+1)-(sqrt(seca-1))(2+√3)]=0 #

So #sqrt(sec a+1)=0#

#=>sec a=-1#

#=>cosa=-1=cospi#

When #0<= a <= 2pi#
#a=pi#

Again

#sqrt(seca+1)-(sqrt(seca-1))(2+√3)=0 #

#=>sqrt((seca+1)/(seca-1))=(2+√3) #

#=>sqrt((1+cosa)/(1-cosa))=(2+√3) #

#=>sqrt((2cos^2(a/2))/(2sin^2(a/2)))=(2+√3) #

#=>tan(a/2)=1/(2+sqrt3)=2-sqrt3=tan(pi/12)#

#=>a/2=pi/12 #

#=>a=pi/6#

Again
#=>tan(a/2)=tan(pi/12)=tan(pi+pi/12)=tan((13pi)/12)#

then #a=(13pi)/6# being #>2pi# the solution neglected.

Note

#tan(pi/12)=tan(pi/4-pi/6)#

#=(tan(pi/4)-tan(pi/6))/(1-tan(pi/4)*tan(pi/6))#

#=(1-1/sqrt3)/(1+1*1/sqrt3)#

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

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

#=(4-2sqrt3)/(3-1)#

#=(4-2sqrt3)/2#

#=2-sqrt3#

Ratnaker Mehta
Feb 2, 2018

#a in {pi/6,pi} sub [0,2pi]#.

Explanation:

We have, #sec^2a-1=tana#.

#:. (seca+1)(seca-1)=tan^2a#.

Sub.ing the given value, #(seca+1)=(2+sqrt3)tana...(1)#, we get,

#(2+sqrt3)tana(seca-1)=tan^2a, or, #

#tan^2a-(2+sqrt3)tana(seca-1)=0#.

#:. tana{tana-(2+sqrt3)(seca-1)}=0#.

#:. tana=0, or, tana=(2+sqrt3)(seca-1)#.

Case 1 : #tana=0#.

# tana=0" with, "a in [0,2pi] rArr a=0,pi,2pi#.

But, #a=0 and 2pi# do not satisfy the given eqn.

#:. a=pi#.

Case 2 : #tana=(2+sqrt3)(seca-1)#.

#tana=(2+sqrt3)(seca-1) rArr seca-1=tana/(2+sqrt3)#.

#:. seca-1=(2-sqrt3)tana............(2)#.

#(1)+(2) rArr 2seca=4tana, i.e., sina=1/2#.

#sina=1/2," read with "a in [0,2pi] rArr a=pi/6, (pi-pi/6)=5pi/6#

Here again, #a=5pi/6# is extraneous.

#:. a=pi/6#.

Altogether, #a in {pi/6,pi} sub [0,2pi]#.

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