Question

if x = tanA - tanB , y = cotB - cotA then 1/x + 1/y is​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x = tanA - tanB$

We know that,

$\boxed{ \rm{ tanx = \frac{1}{cotx}}}$

Using this,

can be rewritten as

$\rm :\longmapsto\:x = \dfrac{1}{cotA} - \dfrac{1}{cotB}$

$\rm :\longmapsto\:x = \dfrac{cotB - cotA}{cotA \: cotB}$

and

$\rm :\longmapsto\:y \: = \: cotB - cotA$

Now,

Consider,

$\rm :\longmapsto\:\dfrac{1}{x} + \dfrac{1}{y}$

On substituting the values of x and y, we get

$\rm \: = \: \: \dfrac{cotAcotB}{cotB - cotA} + \dfrac{1}{cotB - cotA}$

$\rm \: = \: \: \dfrac{cotAcotB + 1}{cotB - cotA}$

$\rm \: = \: \: cotA - cotB$

Hence,

$\bf :\longmapsto\:\dfrac{1}{x} + \dfrac{1}{y} = cotA - cotB$

Additional Information :-

$\boxed{ \rm{ cot(x + y) = \frac{cotxcoty - 1}{coty + cotx}}}$

$\boxed{ \rm{ cot(x - y) = \frac{cotxcoty + 1}{coty - cotx}}}$

$\boxed{ \rm{ tan(x + y) = \frac{tanx + tany}{1 - tanxtany}}}$

$\boxed{ \rm{ tan(x - y) = \frac{tanx - tany}{1 + tanxtany}}}$

$\boxed{ \rm{ sin(x + y) = sinxcosy + sinycosx}}$

$\boxed{ \rm{ sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \rm{ cos(x + y) = cosxcosy - sinxsiny}}$

$\boxed{ \rm{ cos(x - y) = cosxcosy + sinxsiny}}$

$\boxed{ \rm{ {sin}^{2}x - {sin}^{2}y = sin(x + y)sin(x - y)}}$

$\boxed{ \rm{ {cos}^{2}x - {sin}^{2}y = cos(x + y)cos(x - y)}}$