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Question

Mathematics Question on Trigonometric Functions of Sum and Difference of Two Angles

Prove that, tan(π4−x)tan(π4−x)=(1+tanx1−tanx)2.\frac{tan(\frac{π}{4}-x)}{tan(\frac{π}{4}-x)}=(\frac{1+tan x}{1-tan x})^2.

Answer

It is known that

tan(A+B)=tanA+tanB1−tanAtanBtan(A+B)=\frac{tanA+tanB}{1-tanA tanB} and ,

tan(A−B)=tanA−tanB1+tanAtanBtan(A-B)=\frac{tanA-tanB}{1+tanA tanB}

tan(π4+x)tan(π4−x)\frac{tan(\frac{\pi}{4}+x)}{tan(\frac{\pi}{4}-x)}

=(tanπ4+tan x1−tanπ4tan x)(tanπ4−tan x1+tanπ4 tan x)=\frac{(\frac{tan\frac{\pi}{4}+tan\,x}{1-tan\frac{\pi}{4}tan\,x})}{(\frac{tan\frac{\pi}{4}-tan\,x}{1+tan\frac{\pi}{4}\,tan\,x})}

=(1+tan x1−tan x)(1−tan x1+tan x)=\frac{(\frac{1+tan\,x}{1-tan\,x})}{(\frac{1-tan\,x}{1+tan\,x})}

=(1+tan x1−tan x)2=(\frac{1+tan\,x}{1-tan\,x})^2

=R.H.S=R.H.S