Question

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Answer 1

Given:

$\sf x+y+z=\pi$

Prove that:

$\sf cot\dfrac{x}{2}+cot\dfrac{y}{2}+cot\dfrac{z}{2}=cot\dfrac{x}{2}.cot\dfrac{y}{2}.cot\dfrac{z}{2}$

Formula used:

$\boxed{\sf cot\dfrac{\pi}{2}-A=tanA}$

$\boxed{\sf cot\bigg(\dfrac{A}{2} +\dfrac{B}{2}\bigg) =\dfrac{cot\dfrac{A}{2}.cot\dfrac{B}{2}-1}{cot\dfrac{A}{2}+cot\dfrac{B}{2}}}$

Proof:

$\sf According\:to\:the\:question,$

$\sf x+y+z=\pi$

$\Rightarrow\sf x+y=\pi-z$

$\Rightarrow\sf\dfrac{x}{2}+\dfrac{y}{2} =\dfrac{\pi}{2}-\dfrac{z}{2}$

$\sf Taking\:cot\:on\:both\:the\:sides,\:we\:get:$

$\Rightarrow\sf cot\bigg(\dfrac{x}{2}+\dfrac{y}{2}\bigg) =cot\bigg(\dfrac{\pi}{2}-\dfrac{z}{2}\bigg)$

$\Rightarrow \sf \dfrac{cot\dfrac{x}{2}.cot\dfrac{y}{2}-1}{cot\dfrac{x}{2}+cot\dfrac{y}{2}}}=tan\dfrac{z}{2}$

$\Rightarrow \sf \dfrac{cot\dfrac{x}{2}.cot\dfrac{y}{2}-1}{cot\dfrac{x}{2}+cot\dfrac{y}{2}}}=\dfrac{1}{cot\dfrac{z}{2} }$

$\sf On\:Cross\:multiplying,\:we\:get:$

$\Rightarrow \sf {cot\dfrac{x}{2}.cot\dfrac{y}{2}.cot\dfrac{z}{2}-}{cot\dfrac{z}{2}=cot\dfrac{x}{2}+cot\dfrac{y}{2}$

$\Rightarrow \boxed{\sf {cot\dfrac{x}{2}.cot\dfrac{y}{2}.cot\dfrac{z}{2}}=cot\dfrac{x}{2}+cot\dfrac{y}{2}+{cot\dfrac{z}{2}}}}$

Hence proved.

Answer 2

Answer:

$\huge\bold\purple{Given:}$

$\sf x+y+z=\pix+y+z=π$

$\huge{Prove \:that:}$

$\sf cot\dfrac{x}{2}+cot\dfrac{y}{2}+cot\dfrac{z}{2}=cot\dfrac{x}{2}.cot\dfrac{y}{2}.cot\dfrac{z}{2}cot </p><p> </p><p> </p><p></p><p>[tex]\huge{Formula\: used:}$

$\boxed{\sf cot\dfrac{\pi}{2}-A=tanA}$

$\boxed{\sf cot\bigg(\dfrac{A}{2} +\dfrac{B}{2}\bigg) =\dfrac{cot\dfrac{A}{2}.cot\dfrac{B}{2}-1}{cot\dfrac{A}{2}+cot\dfrac{B}{2}}}$

$\huge{Proof:}$

$\sf According\:to\:the\:question,$

$\sf x+y+z=\pix+y+z=π$

$\Rightarrow\sf x+y=\pi-z⇒x+y=π−z$

$\Rightarrow\sf\dfrac{x}{2}+\dfrac{y}{2} =\dfrac{\pi}{2}-\dfrac{z}{2}$

$\sf Taking\:cot\:on\:both\:the\:sides,\:we\:get:$

$\Rightarrow\sf cot\bigg(\dfrac{x}{2}+\dfrac{y}{2}\bigg) =cot\bigg(\dfrac{\pi}{2}-\dfrac{z}{2}\bigg$

$\sf On\:Cross\:multiplying,\:we\:get:$

$\Rightarrow \sf {cot\dfrac{x}{2}.cot\dfrac{y}{2}.cot\dfrac{z}{2}-}$

$\huge\bold\red{Hence \:proved.}$