Question

Hey mate

help me plz​

Answer 1

Hey!

Q If x+y+z=π, prove that:

cot x/2+ cot y/2 +cot z/2=cot x/2 cot y/2 cot z/2

Given: x+y+z=π⇒x2+y2+z2=π2

⇒x2+y2=π2−z2Taking tan on both sides, we gettan(x2+y2)=tan(π2−z2)

⇒tan(x2+y2)=cot(z2)
[∵tan(π2−A)=cotA]

⇒tanx2+tany21−tanx2tany2=cot(z2)
[∵tan(A+B)=tanA+tanB1−tanA tanB]

⇒1cotx2+1coty21-1cotx2×1coty2=cot(z2)

⇒coty2+cotx2cotx2×coty2−1=cot(z2)

⇒coty2+cotx2=cotx2×coty2×cot(z2)−cot(z2)

⇒cotx2+coty2+cot(z2)=cotx2×coty2×cot(z2)

Hope this helps ^_^

Answer 2

Answer:

Step-by-step explanation:

Given $x+y+z=\pi$, prove that:

$\cot\frac{x}{2}+\cot\frac{y}{2}+\cot\frac{z}{2}=\cot\frac{x}{2}\cot\frac{y}{2}\cot\frac{z}{2}$

$x+y+z=\pi\\\\\Rightarrow\frac{x+y+z}{2}=\frac{\pi}{2}\text{ [Divided both side by 2]}\\\\\Rightarrow\frac{x}{2}+\frac{y}{2}+\frac{z}{2}=\frac{\pi}{2}\\\\\Rightarrow\frac{x}{2}+\frac{y}{2}=\frac{\pi}{2}-\frac{z}{2}\\\\\Rightarrow\tan(\frac{x}{2}+\frac{y}{2})=\tan(\frac{\pi}{2}-\frac{z}{2})\text{ [taking tan on both side]}\\\\\Rightarrow\tan(\frac{x}{2}+\frac{y}{2})=\cot\frac{z}{2}\text{ }[\tan(\frac{\pi}{2}-A)=\cot A]$

$\\\\\Rightarrow\frac{\tan\frac{x}{2}+\tan\frac{y}{2}}{1-\tan\frac{x}{2}\tan\frac{y}{2}}=\cot\frac{z}{2}\text{ [we know}\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}]\\\\\Rightarrow\frac{\frac{1}{\cot\frac{x}{2}}+\frac{1}{\cot\frac{y}{2}}}{1-\frac{1}{\cot\frac{x}{2}}\times\frac{1}{\cot\frac{y}{2}}}={\cot\frac{z}{2}}$

$\\\\\Rightarrow\frac{\frac{\cot\frac{y}{2}+\cot\frac{x}{2}}{\cot\frac{x}{2}\times\cot\frac{y}{2}}}{\frac{\cot\frac{x}{2}\times\cot\frac{y}{2}-1}{\cot\frac{x}{2}\times\cot\frac{y}{2}}}=\cot\frac{z}{2}\\\\\Rightarrow\frac{\cot\frac{y}{2}+\cot\frac{x}{2}}{\cot\frac{x}{2}\times\cot\frac{y}{2}-1}=\cot\frac{z}{2}\\\\\Rightarrow\cot\frac{y}{2}+\cot\frac{x}{2}=\cot\frac{x}{2}\times\cot\frac{y}{2}\times\cot\frac{z}{2}-\cot\frac{z}{2}}$

$\\\\\Rightarrow\cot\frac{y}{2}+\cot\frac{x}{2}+\cot\frac{z}{2}} =\cot\frac{x}{2}\times\cot\frac{y}{2}\times\cot\frac{z}{2}\text{ (Proved)}$