How do you prove that tan(x/2)=(secx)/(secxcscx+cscx) ?

2 Answers
Apr 11, 2018

Let #2y =x#. Then the identity becomes

#tan(y) = sec(2y)/(sec(2y)csc(2y) + csc(2y))#

#tany = (1/cos(2y))/(1/(cos(2y)sin(2y)) + 1/sin(2y))#

#tany = (1/cos(2y))/(1/(cos(2y)sin(2y)) + cos(2y)/(cos(2y)sin(2y))#

#tany = ((cos(2y)sin(2y))/cos(2y))/(1 + cos(2y))#

#tany = (sin2y)/(1 + cos(2y))#

#tany = (2sinycosy)/(1 + 2cos^2y - 1)#

#tany = (2sinycosy)/(2cos^2y)#

#tany = siny/cosy#

#LHS = RHS#

As required.

Hopefully this helps!

Apr 11, 2018

Please see below.

Explanation:

We know that,
#color(blue)((1)sectheta=1/costheta and csctheta=1/sintheta#
#color(red)((2)sintheta=2sin(theta/2)cos(theta/2)#
#color(red)((3)1+costheta=2cos^2(theta/2)#
Here,

#tan(x/2)=(secx)/(secxcscx+cscx) #

#RHS=color(blue)((secx)/(secxcscx+cscx).....to Apply(1)#

#=(1/cosx)/(1/(sinxcosx)+1/sinx)#

#=(1/cancelcosx)/((1+cosx)/(sinxcancel(cosx))#

#=1/((1+cosx)/sinx)#

#=color(red)(sinx/(1+cosx)......toApply(2)and (3)#

#=(cancel2sin(x/2)cancelcos(x/2))/(cancel2cos^cancel2(x/2)#

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

#=tan(x/2)#

#=LHS#