Question

#QualityQuestion
@Trigonometry

If $\rm{x + y + z = \pi}$ prove the trigonometric identity;
$\rm{ cot\frac{x}{2} + cot \frac{y}{2} + cotg \frac{z}{2} = cot \frac{x}{2} cot \frac{y}{2} cot \frac{z}{2} }$
​

Answer 1

Step-by-step explanation:

We have,

$x + y + z = \pi$

$\implies \: \frac{x + y + z }{2}= \frac{\pi}{2}$

$\implies \: \frac{x}{2} + \frac{y}{2} + \frac{z}{2} = \frac{\pi}{2}$

$\implies \: \frac{x}{2} + \frac{y}{2} = \frac{\pi}{2} - \frac{z}{2}$

$\implies \: \tan \bigg(\frac{x}{2} + \frac{y}{2} \bigg) = \tan \bigg( \frac{\pi}{2} - \frac{z}{2} \bigg)$

$\implies \: \frac{ \tan( \frac{x}{2} ) + \tan( \frac{y}{2} ) }{1 - \tan( \frac{x}{2} ) \tan( \frac{y}{2} ) } = \cot \bigg( \frac{z}{2} \bigg)$

$\implies \: \frac{ \frac{1}{\cot( \frac{x}{2} ) }+ \frac{1}{ \cot( \frac{y}{2} )} }{1 - \frac{1}{ \cot( \frac{x}{2} ) \cot( \frac{y}{2} ) } } = \cot \bigg( \frac{z}{2} \bigg)$

$\implies \: \frac{ \cot( \frac{x}{2} ) + \cot( \frac{y}{2} ) }{ \cot( \frac{x}{2} ) \cot( \frac{y}{2} ) - 1 } = \cot \bigg( \frac{z}{2} \bigg)$

$\implies \: \cot \bigg( \frac{x}{2} \bigg) + \cot \bigg( \frac{y}{2} \bigg) = \cot \bigg( \frac{x}{2} \bigg) \cot \bigg( \frac{y}{2} \bigg ) \cot \bigg( \frac{z}{2} \bigg) - \cot \bigg( \frac{z}{2} \bigg)$

$\implies \: \cot \bigg( \frac{x}{2} \bigg) + \cot \bigg( \frac{y}{2} \bigg) + \cot \bigg( \frac{z}{2} \bigg) = \cot \bigg( \frac{x}{2} \bigg) \cot \bigg( \frac{y}{2} \bigg ) \cot \bigg( \frac{z}{2} \bigg)$

Answer 2

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

Given that

$\red{\rm :\longmapsto\:x + y + z = \pi \: }$

$\red{\rm :\longmapsto\:x + y = \pi \: - z }$

On dividing each term by 2, we get

$\rm :\longmapsto\:\dfrac{x}{2} + \dfrac{y}{2} = \dfrac{\pi}{2} - \dfrac{z}{2}$

So,

$\rm :\longmapsto\: cot\bigg(\dfrac{x}{2} + \dfrac{y}{2} \bigg) =cot\bigg( \dfrac{\pi}{2} - \dfrac{z}{2}\bigg)$

We know that,

$\boxed{ \sf{cot\bigg(\dfrac{\pi}{2} - x \bigg) = tanx}}$

and

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

So, using these Identities, we get

$\rm :\longmapsto\:\dfrac{cot\bigg(\dfrac{x}{2} \bigg)cot\bigg(\dfrac{y}{2} \bigg) - 1 }{cot\bigg(\dfrac{x}{2} \bigg) + cot\bigg(\dfrac{y}{2} \bigg)} = tan\bigg(\dfrac{z}{2} \bigg)$

$\rm :\longmapsto\:\dfrac{cot\bigg(\dfrac{x}{2} \bigg)cot\bigg(\dfrac{y}{2} \bigg) - 1 }{cot\bigg(\dfrac{x}{2} \bigg) + cot\bigg(\dfrac{y}{2} \bigg)} = \dfrac{1}{cot\bigg(\dfrac{z}{2} \bigg)}$

On cross multiplication, we get

$\rm :\longmapsto\:cot\bigg(\dfrac{x}{2} \bigg)cot\bigg(\dfrac{y}{2} \bigg)cot\bigg(\dfrac{z}{2} \bigg) - cot\bigg(\dfrac{z}{2} \bigg) = cot\bigg(\dfrac{x}{2} \bigg) + cot\bigg(\dfrac{y}{2} \bigg)$

On transposition, we get

$\rm :\longmapsto\:cot\bigg(\dfrac{x}{2} \bigg)cot\bigg(\dfrac{y}{2} \bigg)cot\bigg(\dfrac{z}{2} \bigg) = cot\bigg(\dfrac{z}{2} \bigg) + cot\bigg(\dfrac{x}{2} \bigg) + cot\bigg(\dfrac{y}{2} \bigg)$

can be rewritten as

$\rm :\longmapsto\:cot\bigg(\dfrac{x}{2} \bigg) + cot\bigg(\dfrac{y}{2} \bigg) + cot\bigg(\dfrac{z}{2} \bigg) = cot\bigg(\dfrac{x}{2} \bigg)cot\bigg(\dfrac{y}{2} \bigg)cot\bigg(\dfrac{z}{2} \bigg)$

Hence, Proved

Additional Information :-

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

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

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

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

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

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

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

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