[tex]if \: x+y+z=\pi[/tex] prove the trigonometric identity
[tex]cot{\frac{x}{2}}+cot{\frac{y}{2}}+cotg\frac{z}{2}=cot{\frac{x}{2}}cot{\frac{y}{2}}cot{\frac{z}{2}}[/tex]
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[tex]cot{\frac{x}{2}}+cot{\frac{y}{2}}+cotg\frac{z}{2}=cot{\frac{x}{2}}cot{\frac{y}{2}}cot{\frac{z}{2}}[/tex]
Only above than genius allowed to answer
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Verified answer
Solution :
Given, x + y + z = π
To prove : [tex]\displaystyle\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} }[/tex]
Now, as [tex]\sf{ x + y + z = 180° = \pi }[/tex]
Now dividing both sides by 2,
[tex]\longrightarrow\sf{ \dfrac{x + y + z}{2} = \dfrac{ \pi }{2} }[/tex]
[tex]\longrightarrow\sf{ \dfrac{x}{2} + \dfrac{y}{2} + \dfrac{z}{2} = \dfrac{ \pi }{2} }[/tex]
[tex]\longrightarrow\sf{ \dfrac{x}{2} + \dfrac{y}{2} = \dfrac{\pi}{2} - \dfrac{z}{2} }[/tex]
[tex]\longrightarrow\sf{ cot \bigg( \dfrac{x}{2} + \dfrac{y}{2} \bigg) = cot \bigg(\dfrac{\pi}{2} - \dfrac{z}{2} \bigg) }[/tex]
[tex]\boxed{\bold{Formula : cot(A + B) = \dfrac{cotAcotB - 1}{cotB + cotA} }}[/tex]
[tex]\longrightarrow\sf{ \dfrac{cot \dfrac{x}{2} cot \dfrac{y}{2} - 1}{cot \dfrac{y}{2} + cot \dfrac{x}{2} } = tan \dfrac{z}{2} \: \bigg[\because cot \bigg(\dfrac{\pi}{2} - \theta \bigg) = tan \theta \bigg] }[/tex]
[tex]\longrightarrow\sf{ \dfrac{cot \dfrac{x}{2} cot \dfrac{y}{2} - 1}{cot \dfrac{y}{2} + cot \dfrac{x}{2} } = \dfrac{1}{ cot \dfrac{z}{2}} }[/tex]
[tex]\longrightarrow\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} }[/tex]
[tex]\longrightarrow\underline{\boxed{\bold{\blue{ cot \dfrac{x}{2} + cot \dfrac{y}{2} + cot \dfrac{z}{2} = cot \dfrac{x}{2} cot \dfrac{y}{2} cot \dfrac{z}{2} }}}}[/tex]
Hence, proved!