Question

\[\frac{ \tan x+\tan y }{ \cot x+\cot y }=\tan x \tan y\]

Answer 1

\(\dfrac{ \tan x+\tan y }{ \cot x+\cot y }=\tan x \tan y\)

\(\dfrac{ \dfrac{\sin x}{\cos x}+\dfrac{\sin y}{\cos y} }{ \dfrac{\cos x}{\sin x}+\dfrac{\cos y}{\sin y} }=\tan x \tan y\)

\(\dfrac{ \dfrac{\sin x}{\cos x} \dfrac{\cos y}{\cos y} +\dfrac{\sin y}{\cos y} \dfrac{\cos x}{\cos x}}{ \dfrac{\cos x}{\sin x} \dfrac{\sin y}{\sin y}+\dfrac{\cos y}{\sin y} \dfrac{\sin x}{\sin x}}=\tan x \tan y\)

\(\dfrac { \dfrac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} } { \dfrac{\cos x \sin y + \sin x \cos y}{\sin x \sin y}} =\tan x \tan y\)

\( \dfrac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} \div { \dfrac{\cos x \sin y + \sin x \cos y}{\sin x \sin y}} =\tan x \tan y\)

\( \dfrac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} \times \dfrac {\sin x \sin y} {\cos x \sin y + \sin x \cos y} =\tan x \tan y\)

\( \dfrac{\cancel{\sin x \cos y + \cos x \sin y}}{\cos x \cos y} \times \dfrac {\sin x \sin y} {\cancel{\cos x \sin y + \sin x \cos y}} =\tan x \tan y\)

\( \dfrac{\sin x \sin y} {\cos x \cos y} =\tan x \tan y\)

\(\tan x \tan y = \tan x \tan y \)

Answer 2

@mathstudent55 Thank you!

Answer 3

wlcm

Answer 4

you know cot = \(\dfrac{1}{tan}\) there fore, your is