Question

if A + B = 225° then (1 + cot A)/cot A × (1 + cot B)/cot B = ?​

Answer 1

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

Given that

$\rm :\longmapsto\:A + B = 225 \degree$

can be rewritten as

$\rm :\longmapsto\:tan(A + B )= tan225 \degree$

$\rm :\longmapsto\:\dfrac{tanA + tanB}{1 - tanAtanB} = tan(90 \degree \times 2 + 45 \degree)$

$\rm :\longmapsto\:\dfrac{tanA + tanB}{1 - tanAtanB} = tan 45 \degree$

$\rm :\longmapsto\:\dfrac{tanA + tanB}{1 - tanAtanB} = 1$

$\rm :\longmapsto\:tanA + tanB = 1 - tanAtanB$

$\rm :\longmapsto\:tanA + tanB + tanAtanB = 1$

On adding 1 on both sides, we get

$\rm :\longmapsto\:1 + tanA + tanB + tanAtanB = 1 + 1$

$\rm :\longmapsto\:(1 + tanA) + tanB(1+ tanA) = 2$

$\rm :\longmapsto\:(1 + tanA)(1+ tanB) = 2$

$\rm :\longmapsto\:\bigg(1 + \dfrac{1}{cotA} \bigg) \bigg(1 + \dfrac{1}{cotB} \bigg) = 2$

$\bf :\longmapsto\:\bigg(\dfrac{cotA + 1}{cotA} \bigg) \bigg(\dfrac{cotB + 1}{cotB} \bigg) = 2$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

$\underbrace{\boxed{ \tt{tan(180 \degree + x) = tanx}}}$

$\underbrace{\boxed{ \tt{tanx = \frac{1}{cotx} }}}$

Additional Information :-

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

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

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

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

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