Tan (pie/4+x)/tan (pie/4-x)={(1+tan x)/(1-tan x)}^2??

3 Answers
Narad T.
May 21, 2018

Please see the proof below

Explanation:

We need

#tan(a+b)=(tana+tanb)/(1-tanatanb)#

#tan(a-b)=(tana-tanb)/(1+tanatanb)#

#tan(pi/4)=1#

Therefore,

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

#=((tan(pi/4)+tanx)/(1-tan(pi/4)tanx))/((tan(pi/4)-tanx)/(1+tan(pi/4)tanx))#

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

#=(1+tanx)^2/(1-tanx)^2#

#=((1+tanx)/(1-tanx))^2#

#=RHS#

#QED#

Ratnaker Mehta
May 21, 2018

Kindle refer to Explanation for a Proof.

Explanation:

Let, #tanx=t#.

Then, #tan(pi/4+x)={tan(pi/4)+tanx}/{1-tan(pi/4)tanx}#,

#:. tan(pi/4+x)=(1+t)/(1-t).............(ast^1)#.

Likewise, #tan(pi/4-x)=(1-t)/(1+t)............(ast^2)#.

Hence, by #(ast^1) and (ast^2)#, we have,

#tan(pi/4+x)/tan(pi/4-x)=(1+t)/(1-t)-:(1-t)/(1+t)#,

#={(1+t)/(1-t)}^2#,

#={(1+tanx)/(1-tanx)}^2#, as desired!

Ratnaker Mehta
May 21, 2018

Please refer to Explanation for another Proof.

Explanation:

We know that, #tan(pi/2-theta)=cottheta=1/tantheta#.

Replacing #theta# bay #(pi/4-x)#, we, from this, get,

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

Utilising this, we have, #{tan(pi/4+x)/tan(pi/4-x)}#,

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

#=tan(pi/4+x)*tan(pi/4+x)#,

#={tan(pi/4+x)}^2#,

#={(tan(pi/4)+tanx)/(1-tan(pi/4)*tanx)}^2#,

#={(1+tanx)/(1-tanx)}^2#, as before!

Enjoy Maths.!

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