Question

if x+y -z= π ,prove the trigonometric identity cotx/2 + coty/2 + cot z/2= cotx/2.coty/2.cotz/2

Answer 1

there is an error in the question..the question must be:; if x+y+z= pi....then prove...

Answer 2

Answer:  The proof is done below.

Step-by-step explanation:  We are given that :

$x+y+z=\pi~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to prove that

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

We ill be using the following formula :

$(i)~\cot(A+B)=\dfrac{\cot A\cot B-1}{\cot A+\cot B},\\\\\\(ii)~\cot(\dfrac{\pi}{2}+A)=-\tan A,\\\\\\(iii)~\cot A=\dfrac{1}{\tan A}.$

From equation (i), we have

$x+y+z=\pi\\\\\Rightarrow \dfrac{x+y+z}{2}=\dfrac{\pi}{2}~~~~~~~~~~~~~~~~~~~[\textup{dividing both sides by 2}]\\\\\Rightarrow\dfrac{x}{2}+\dfrac{y}{2}=\dfrac{\pi}{2}-\dfrac{z}{2}\\\\\Rightarrow \cot\left(\dfrac{x}{2}+\dfrac{y}{2}\right)=\cot\left(\dfrac{\pi}{2}-\dfrac{z}{2}\right)\\\\\\\Rightarrow \dfrac{\cot\dfrac{x}{2}\cot\dfrac{y}{2}-1}{\cot\dfrac{x}{2}+\cot\dfrac{y}{2}}=\tan\dfrac{z}{2}\\\\\\\Rightarrow \dfrac{\cot\dfrac{x}{2}\cot\dfrac{y}{2}-1}{\cot\dfrac{x}{2}+\cot\dfrac{y}{2}}==\dfrac{1}{\cot\dfrac{z}{2}}\\\\\\\Rightarrow \cot\dfrac{x}{2}\cot\dfrac{y}{2}\cot\dfrac{z}{2}-\cot\dfrac{z}{2}=\cot\dfrac{x}{2}+\cot\dfrac{y}{2}\\\\\Rightarrow \cot\dfrac{x}{2}+\cot\dfrac{y}{2}+\cot\dfrac{z}{2}=\cot\dfrac{x}{2}\cot\dfrac{y}{2}\cot\dfrac{z}{2}.$

Hence proved.