Question

Prove the following:
$\dfrac{{\tan \left( {\dfrac{\pi }{4} + x} \right)}}{{\tan \left( {\dfrac{\pi }{4} - x} \right)}} = {\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)^2}$

Answer 1

Hint: Note that, $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ and $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
Use these formulas to simplify the left hand side and proceed.
On simplification we will get our desired result.

Complete step-by-step answer:
Given to prove that $\dfrac{{\tan \left( {\dfrac{\pi }{4} + x} \right)}}{{\tan \left( {\dfrac{\pi }{4} - x} \right)}} = {\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)^2}$,
Left hand side is given by:
$ = \dfrac{{\tan \left( {\dfrac{\pi }{4} + x} \right)}}{{\tan \left( {\dfrac{\pi }{4} - x} \right)}}$
Using, $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ and $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$, we get,
$ = \dfrac{{\dfrac{{\tan \dfrac{\pi }{4} + \tan x}}{{1 - \tan \dfrac{\pi }{4} \times \tan x}}}}{{\dfrac{{\tan \dfrac{\pi }{4} - \tan x}}{{1 + \tan \dfrac{\pi }{4} \times \tan x}}}}$
As, ${\text{tan}}\dfrac{\pi }{4}{\text{ = 1}}$, we get,
$ = \dfrac{{\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)}}{{\left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)}}$
On simplification we get,
$ = \dfrac{{1 + \tan x}}{{1 - \tan x}} \times \dfrac{{1 + \tan x}}{{1 - \tan x}}$
As we can club the common terms, so we get,
$ = {\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)^2}$
= Right hand side
Hence, $\dfrac{{\tan \left( {\dfrac{\pi }{4} + x} \right)}}{{\tan \left( {\dfrac{\pi }{4} - x} \right)}} = {\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)^2}$
(proved)

Note: Note the following important formulae,
1.$\sin A\cos B + \cos A\sin B = \sin \left( {A + B} \right)$
2.$\sin A\cos B - \cos A\sin B = \sin \left( {A - B} \right)$
3.$\cos A\cos B - \sin A\sin B = \cos \left( {A + B} \right)$
4.$\cos A\cos B + \sin A\sin B = \cos \left( {A - B} \right)$
5.$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
6.$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
Also, the trigonometric ratios of the standard angles are given by

	\[0^\circ \]	\[30^\circ \]	\[45^\circ \]	\[60^\circ \]	\[90^\circ \]
\[\operatorname{Sin} x\]	0	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	1
\[\operatorname{Cos} x\]	1	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	0
\[\operatorname{Tan} x\]	0	$\dfrac{1}{{\sqrt 3 }}$	1	$\sqrt 3 $	Undefined
\[Cotx\]	undefined	$\sqrt 3 $	1	$\dfrac{1}{{\sqrt 3 }}$	0
\[\cos ecx\]	undefined	2	$\sqrt 2 $	$\dfrac{2}{{\sqrt 3 }}$	1
\[\operatorname{Sec} x\]	1	$\dfrac{2}{{\sqrt 3 }}$	$\sqrt 2 $	2	Undefined