How do you prove that $\frac { \cos 7x + \cos 5x } { \sin 7x - \sin 5x } = \cot x$ ?

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Answer 1 · 2017-11-27T19:35:22

See the proof below

We need

$cosa+cosb=2cos((a+b)/2)cos((a-b)/2)$

$sina-sinb=2cos((a+b)/2)sin((a-b)/2)$

$cotx=cosx/sinx$

Here,

$a=7x$ and $b=5x$

Therefore,

$LHS=(cos7x+cos5x)/(sin7x-sin5x)=(2cos((7x+5x)/2)cos((7x-5x)/2))/(2cos((7x+5x)/2)sin((7x-5x)/2))$

$=(cos(6x)cos(x))/(cos(6x)sin(x))$

$=cosx/sinx$

$=cotx$

$=RHS$

$QED$

How do you prove that \frac { \cos 7x + \cos 5x } { \sin 7x - \sin 5x } = \cot x\frac { \cos 7x + \cos 5x } { \sin 7x - \sin 5x } = \cot x?