Question #e51b8

3 Answers
Jan 22, 2017

#tan (x/2 + pi/4) = sec x + tan x#

Explanation:

Use the trig identity:
#tan (a + b) = (tan a + tan b)/(1 - tan a.tan b)#
Call t = (x/2) and develop the left side
#LS = tan (t + pi/4) = (tan t + tan (pi/4))/(1 - tan t.tan (pi/4))#
Trig table gives #tan (pi/4) = 1#, there for:
#LS = (1 + tan t)/(1 - tan t) = ((cos t + sin t)/(cos t))(cos t/(cos t - sin t)) =#
#LS = (cos t + sin t)/(cos t - sin t)#
Multiply both numerator and denominator by (cos t + sin t), we get:
#LS = (cos t + sin t)^2/(cos^2 t - sin^2 t)#
Reminder:
#(cos t + sin t)^2 = 1 + 2cos t.sin t = 1 + sin 2t#
#cos^2 t - sin^2 t = cos 2t#.
#LS = (1 + sin 2t)/(cos 2t) = 1/(cos 2t) + (sin 2t)/(cos 2t)#
#LS = sec 2t + tan 2t#
Replace t by #(x/2)#, we get
#tan (x/2 + pi/4) = sec x + tan x#

Jan 22, 2017

#RHS=secx+tanx#

#=1/cosx+sinx/cosx#

#=(1+sinx)/cosx#

#=(cos^2(x/2)+sin^2(x/2)+2sin(x/2)cos(x/2))/(cos^2(x/2)-sin^2(x/2))#

#=(cos(x/2)+sin(x/2))^2/((cos(x/2)+sin(x/2))(cos(x/2)-sin(x/2)))#

#=(cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2))#

#=(cos(x/2)/cos(x/2)+sin(x/2)/cos(x/2))/(cos(x/2)/cos(x/2)-sin(x/2)/cos(x/2))#

#=(1+tan(x/2))/(1-tan(x/2))#

#=(tan(pi/4)+tan(x/2))/(1-tan(x/2)tan(pi/4))#

#=tan(x/2+pi/4)=LHS#

Jan 22, 2017

Proof given below

Explanation:

#tan (x/2 +pi/4)= (tan pi/4 + tan x/2)/(1- tan x/2 tan pi/4)#

=#(1+ tan x/2)/(1-tan x/2)# = #(cos x/2 +sinx/2)/(cos x/2- sin x/2)#

Now multiply the numerator and denominator by #(cos x/2 + sin x/2)#

= #(1+ 2sin x/2 cos x/2)/(cos^2 x/2 - sin^2 x/2)# = #(1+sin x)/cos x#= sec x + tanx